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vedanta Excel In Opt. Mathematics - Book 10 (VIII) cosD - cosC = 2sin C + D. sin C-D . 2 2 C + D. D-C or, cosC - cosD = 2 sin 2 sin 2 Above formulae (i) to (viii) are the transformation formulae of trigonometric ratios. Worked out Examples Example 1. Express the following product forms into sum or difference forms: Solution: (a) sin24° cos12° (b) cos42° sin22° Example 2. Solution: (c) cos50°.cos35° (d) sin40° sin20° (a) sin24°.cos12° = 1 (2sin24°cos12°) 2 1 = 2 [sin(24° + 12°) + sin(24° - 12°)] = 1 [sin36° + sin12°] 2 1 (b) cos42°. sin22° = 2 [2cos42°.sin22°] = 1 [sin(42° + 22°) - sin(42° - 22°)] 2 1 = 2 [sin64° - sin20°] (c) cos50° cos35° = 1 [2cos50° cos35°] 2 1 = 2 [cos(50° + 35°) + cos(55° - 35°)] = 1 [cos85° + cos20°] 2 1 (d) sin40°. sin20° = 2 [2sin40°.sin20°] 1 [cos(40° - 20°)- cos(40° + 20°)] (= 2 1 1 1 = 2 [cos20° - cos60°] = 2 cos20°- 2 ( Express the following sum or difference into product form: (a) cos6T + cos4T (b) sin50° - sin40° (c) sin50° - sin40° (d) cos35° - cos25° (a) cos6T + cos4T = 2cos 6T + 4T . cos 6T - 4T 2 2 = 2cos5T. cosT (b) sin50° + sin40° = 2sin 50° + 40° . cos 50° - 40° 2 2 = 2sin45°.cos5° (c) sin50° - sin40° = 2cos 50° + 40° . sin 50° - 40° 2 2 = 2cos45°. sin5° Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 225

vedanta Excel In Opt. Mathematics - Book 10 (d) cos35° - cos25° = 2sin 35° + 25° . sin 25° - 35° 2 2 = 2sin30°.sin(-5°) = –2sin30°.sin5° [ sin(-T) = - sinT@ = –2 . 1 . sin15° = – sin15° 2 Example 3. Prove that Solutions: (a) sin75° + sin15° = 3 (b) sin18° + cos18° = 2 cos27° 2 1 (c) sin(45° - A) sin (45° + A) = 2 cos2A (a) LHS = sin75° + sin15° = 2sin 75° + 15° . cos 75° - 15° 2 2 = 2sin45°. cos30° =2. 1 . 3 = 3 = RHS Proved. 2 2 2 (b) LHS = sin 18° + cos18° = sin18° + cos(90° - 72°) = sin18° + sin72° ( ) ( )= 2sin 18° + 72° . cos 72° - 18° 2 2 1 = 2sin45°. cos27° = 2. 2 . cos27° = 2 cos27° = RHS. Proved. (c) LHS = sin (45° - A). sin (45° + A) = 1 [2sin(45° - A). sin(45° + A)] 2 1 = 2 [cos(45° - A - 45° - A) - cos(45° - A + 45° + A)] = 1 [cos (-2A)- cos90°] = 1 cos2A = RHS. Proved. 2 2 Example 4. Proved that Solution: (a) sin4A + sin2A = tan3A (b) cos8° + sin8° = tan53° cos4A + cos2A cos8° – sin8° (a) LHS = sin4A + sin2A = tan3A cos4A + cos2A ( ) ( )2sin 4A + 2A . cos 4A - 2A = sin3A . cosA = 2 . cos 2 cos3A. cosA ( ) ( )2cos4A + 2A 4A - 2A 2 2 = tan3A = RHS Proved (b) LHS = cos8° + sin8° cos8° – sin8° cos8° + sin(90° - 82°) cos8° + cos82° = cos8° - sin(90° - 82°) = cos8° - cos82° 226 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (( )) (( ))=2cos8 + 82° . cos 8°-82° = sin45°. cos37° 2sin 2 . sin 2 sin45°.sin37° cos37° 8° + 82° 82° - 8 sin37° 2 2 = = cot 37° = cot (90° - 53°) = tan53° = RHS. Proved. Example 5. Proved that sin7T - sin5T - sin3T + sinT = tan2T Solution: cos7T + cos3T - cos5T - cosT Example 6. LHS = sin7T - sin5T - sin3T + sinT Solution: cos7T + cos3T - cos5T - cosT Example 7. = (sin7T + sinT) - (sin5T + sin3T) Solution: (cos7T - cosT) + (cos3T - cos5T) 2sin 7T + T . cos 7T – T – 2sin 3T + 5T . cos 5T – 3T 2sin 2 2 2 2 = 7T + T T – 7T 3T + 5T 5T – 3T 2 . sin 2 + 2sin 2 . sin 2 = sin4T . cos3T - sin4T . cosT = sin4T (cos3T - cosT) sin4T . sin(–3T) + sin4T .sinT sin4T (sinT - sin3T) 2sin 3T + T . sin T - 3T 2 2 sin2T = 2cos T + 3T . sin T - 3T = cos2T = tan2T = RHS. Proved. 2 2 Prove that sin8T . cosT - sin6T . cos3T = tan2T cos2T . cosT - sin3T . sin4T sin8T . cosT - sin6T . cos3T LHS = cos2T . cosT - sin3T . sin4T = 2sin8T . cosT - 2sin6T . cos3T 2cos2T . cosT - 2sin3T . sin4T = sin(8T + T) + sin(8T - T) - sin(6T + 3T) - sin(6T - 3T) cos(2T + T) + cos(2T - T) - cos(3T - 4T) + cos(3T + 4T) = sin9T + sin7T- sin9T - sin3T = sin7T - sin3T cos3T + cosT - cosT + cos7T cos3T + cos7T = 2cos 7T +3T . sin T - 3T = sin2T 2 2 cos2T 2cos T + 7T . cos 3T - 7T 2 2 = tan2T = RHS. Proved. Prove that sin20°.sin40°.sin60°.sin80°.Sun120° = 33 32 LHS = sin20°.sin40°.sin60°.sin80°.sin120° = sin20° sin40° 3 sin80° 3 2 2 3 = 4×2 sin20° [2sin40°. sin80°] Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 227

vedanta Excel In Opt. Mathematics - Book 10 = 3 . sin20° [cos(40° - 80°) - cos(40° + 80°)] 8 3 = 8 sin20° [cos40° - cos120°] = 3 sin20° .cos40° - 3 sin20° (– 1 ) 8 8 2 3 3 = 8 sin20° .cos40° + 16 sin20° = 3 [sin(20° + 40°) + sin(20° - 40°)] + 3 sin20° 16 16 3 3 = 16 [sin60° + sin(-20°)] + 16 sin20° = 3 . 3 - 3 sin20° + 3 sin20° = 33 = RHS. Proved. 16 2 16 16 32 1 Example 8. Prove that cos20°.cos40°. cos60°.cos80° = 16 . Solution: LHS = cos20°. cos40°.cos60°.cos80° Example 9. Solution: = cos20°.cos40° 1 . cos80° 2 228 1 = 4 cos20° (2cos40°.cos80°) = 1 cos20° [cos(40° + 80°) + cos(40° - 80°)] 4 1 (= 4 cos20° [cos120° + cos40°] = 1 cos20° - (+ 1 cos20°.cos40° 4  ( 4 1 1 = – 8 cos20° + 8 (2cos20° . cos40°) = – 1 cos20° + 1 [cos(20° + 40°) + cos(20° - 40)] 8 8 1 1 1 1 1 1 = – 8 cos20° + 8 cos60° + 8 cos20° = 8 . 2 = 16 = RHS. Proved. Alternative Method LHS = cos20°. cos40°.cos60°.cos80° = 1 (2 sin20°cos20° ) cos40° 1 cos80° 2sin20° 2 1 = 4sin20° sin40°cos40° cos80° = 1 (2sin40°cos40°).cos80° = 1 (2sin80°cos80°) 8sin20° 16sin20° 1 1 1 = 16sin20° sin160° = 16sin20° . sin20° = 16 = RHS Proved. Prove that sinT.sin(60° - T) sin(60° + T) = 1 sin3T 4 LHS = sinT.sin(60° - T) sin(60° + T) = 1 sinT [2sin(60° - T). sin (60° + T)] 2 1 = 2 sinT [cos(60° - T - 60 - T) - cos(60° - T + 60° + T)] (= 1 [cos 2 sinT [-2T] - cos120°] = 1 sinT.cos2T - 1 . sinT - 2 2  Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = 1 [sin (T + 2T) + sin (T - 2T)] + 1 sinT 4 4 1 1 1 1 = 4 sin3T - 4 sinT + 4 sinT = 4 sin3T = RHS. Proved. Example 10. Prove that cos Sc . cos 2Sc . cos 4Sc . cos 7Sc = 1 Solution: 15 15 15 15 16 Sc 2Sc 4Sc 7Sc Example 11. (LHS=cos 15 , cos 15 , cos 15 . cos 15 Solution: Sc Sc cos 2Sc . cos 4Sc . cos 7Sc 15     2sin .cos (( (( = 2sin Sc ( 15 = sin 2Sc . cos 2Sc ( . cos 4Sc . cos 7Sc (= 15 15 ( 15 15 2sin 2Sc .cos (2Sc . cos 4Sc . cos 7Sc 15 (   ( 4sin Sc (= 15 ( 1 2sin 4Sc .cos 4Sc . cos 7Sc 8sin Sc 15   (= 15 1 sin 8Sc . cos cos 7Sc 8sin Sc   (= 15 1 sin Sc - 7Sc .cos 7Sc 8sin Sc   (= 15 1 2sin 7Sc cos 7Sc 16sin Sc   15 1 16sin sin 14Sc (= Sc  15 1 = 16sin Sc . sin Sc - Sc 15  = 1 . sin Sc = 1 = RHS. Proved.   16sin Sc 15 Prove that 2cos Sc . cos91S3c + cos31S3c + cos51S3c = 0 13 Sc cos91S3c 3Sc cos51S3c 2cos 13 13 + ( (LHS= . + cos = cos Sc + 9Sc + cos Sc – 9Sc + 3Sc + cos 5Sc 13 13 13 13 13 13 10Sc 8Sc 3Sc 5Sc 13 13 13 13 ( (= cos + cos + cos + cos = cos Sc – 3Sc + cos Sc – 5Sc + cos 3Sc + cos 3Sc 13 13 13 13 3Sc 5Sc 3Sc 5Sc = - cos 13 - cos 13 + cos 13 + cos 13 = 0 = RHS Proved. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 229

vedanta Excel In Opt. Mathematics - Book 10 Example 12. (HIfesrei1n, Dsi1n+Dc+o1sDco=1sDsi=1nEs+i1nEco1+sEprove that cot D+E (, = tanD.tanE. Solution: 1 2 cosE 1 1 1 1 or, sinD - sinE = cosE - cosD or, sinE- sinD = cosD- cosE sinD.sinE cosD.cosE + E-D E-D 2cos E 2 D . sin 2 2sin DE . sin 2 2 or, = ((or, sinD . sinE cosD . cosE cos D+E ( sinD. sinE 2 ( cosD.cosE = D+E ( sin 2 ( (? ( cot D+E = tanD.tanE Proved. 2 Example 13. Prove that sin2D - sin2E = tan(D + E) sinD.cosD - sinE.cosE sin2D - sin2E 2sin2D - 2sin2E Solution: LHS = sinD.cosD - sinE.cosE = 2sinD.cosD - 2sinE.cosE 1-cos2D - 1 + cos2E cos2E - cos2D sin2D - sin2E sin2D - sin2E ( (= = 2sin 2E + 2D . sin 2D - 2E ( ( (= 2 2 ( sin(D + E) 2cos 2D + 2E . sin 2D - 2E = cos(D + E) 22 = tan (D + E) = RHS. Proved. Example 14. If sin2x + sin2y = 1 and cos2x + cos2y = 1 , then show that 2 3 2 tan (x + y) = 3 . Solution: ( (Here, sin2x + sin2y = 1 3 1 2x + 2y 2x - 2y 3 2 or, 2sin (( . cos 2 = or, (( ............. (i) 2sin (x + y).cos (x - y) = 1 3 1 ( (Again, cos2x + cos2y = 2 1 2 2cos 2x + 2y . cos 2x - 2y = or, 2 + y). cos(x y) =221 - ............. (ii) 2cos(x Dividing (i) by (ii), we get, 2sin(x + y). cos(x - y) 1 2cos(x + y).cos(x - y) 3 = 1 tan (x + y) = 2 Proved. 2 3 ? 230 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Exercise 9.3 Very Short Questions 1. Express each of the following as sum or difference form: (a) 2sinA. sinB (b) 2cosA.cosB (c) 2cosA.sinB (d) 2sinA.cosB (e) 2sin5T. cos3T (f) 2sin3x. sinx (g) 2sin20°.sin10° (h) 2cos40°.cos20° 2. Express each of the following as product form: (a) sin6T + sin2T (b) sinx - siny (c) cosx - cosy (d) sinx + siny (e) sin50° + sin40° (f) cos70° - cos40° (g) cos40° - sin20° (h) sin5x - sin3x (i) cos7x - cos11x (j) cos5x + cos2x 3. Prove the following: (a) cos75° + cos15° = 3 (b) sin75° - sin15° = 1 2 2 (d) sin50° + sin70° = 3 cos10° (c) sin18° + cos18° = 2 cos27° (e) sin50° + sin10° = sin70° (f) cos52° + cos68° + cos172° = 0 Short Questions 4. Prove the following : (a) cos5A + cos3A = cotA (b) cos40° - cos60° = tan50° sin5A - sin3A sin60° - sin40° cos80° + cos20° cos8° + sin8° (c) sin80° - sin20° = 3 (d) cos8° - sin8° = tan53° (e) cos10° - sin10° = cot55° (f) cos(40° + A) + cos(40° - A) = cotA cos10°+ sin10° sin(40° + A) - sin(40° - A) Long Questions 5. Prove the following : (a) sinA.sin2A + sin3A.sin6A = tan5A sinA.cos2A + sin3A.cos6A cos2A.cos3A - cos2A.cos7A sin7A + sin3A (b) sin4A.sin3A - sin2A.sin5A = sinA (c) sinA + sin3A + sin5A + sin7A = tan4A cosA + cos3A + cos5A + cos7A cos7A + cos3A - cos5A -cosA (d) sin7A - sin3A - sin5A + sinA = cot2A (e) sin5A - sin7A - sin4A + sin8A = cot6A cos4A - cos5A - cos8A + cos7A sin(p+2)T - sinpT (f) cospT - cos(p + 2) T = cot (p + 1) T (g) (sin4A + sin2A) (cos4A - cos8A) =1 (sin7A + sin5A) (cosA - cos5A) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 231

vedanta Excel In Opt. Mathematics - Book 10 6. Prove the following : (a) ssccsiiioonnnss2124200000°°°°°.....cssscoiiionnnss4434100000°°°°0....css°siio.inncsno8665s00001°°°°.6.=.sc0ison°isn8=8837000°°18°===136111616 (b) (c) (d) (e) (f) tan20° tan40° tan80° = 3 7. Prove the following : (a) sin(45° + T). sin(45° - T) = 1 cos2T 2 1 (b) cos(45° + T). cos(45° - T) = 2 cos2T (c) cosT. cos(60° – T) . cos(60 +T) = 1 cos3T 4 1 (d) sinx. sin(60° – x). sin(60° + x) = 4 sin3x 8. Prove that cos(36° - T).cos(36° + T) + cos(54° + T). cos(54° - T) = cos2T ((a) 9. Prove the following : If 1 – 1 =- 1 – co1sB, prove that cot A+B ( = – tanA.tanB sinA cosA sinB 2 ( ((b) If 4 + 2 = 2 + 4 , prove that tan D+T = 1 secT secD cosecD secD 2 2 ((a) 10. Prove the following : A-B ( 2 ( (cosA + cosB)2 + (sinA + sinB)2 = 4cos2 ((b) ( ((c) = 4sin2 A -B 2 (d) 3 (cosB - cosA)2 + (sinA - sinB)2 2 sin2 sin2A Sc + A (( Sc A 1 sinA 8 2 - sin2 8 - 2 = 2 + sin2 (A + 120°) + sin2 (A - 120°) = (11. (a) 1 1 D+E 1 4 2 2 2 If sinD+ sinE= and cosD+cosE = , then(( prove that : tan = ( ( ((b) 1 1 D+E 3 If cosD + cosE = 3 and sinD + sinE = 4 , prove that : tan 2 = 4 ( ((c) If sinx = k siny, prove that : tan x-y = k-1 tan x+y 2 k +1 2 (d) If sin (A + B) = k sin (A - B), then prove that : (k - 1) tanA = (k + 1) tanB. 12. Prove that sin2x - sin2y = tan(x + y) sinx.cosx - siny.cosy 1 (13. Prove that cos 2Sc + cos 4Sc + cos 6Sc = - 2 7 7 7 D+E ( 14. Prove that x = y cot 2 if xcosD + ysinD = xcosE + ysinE 15. (a) Prove that 1 – cos10° + cos40° – cos50° = tan5°.cot20° 1 + cos10° – cos40° – cos50° 232 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (b) Prove that 1 – cosT + cosD – cos(DT) = tan T .cot D 1 + cosT – cosD – cos(D+T) 2 2 1. (a) cos(A - B) - cos(A + B) (b) cos(A + B) + cos(A - B) (c) sin(A + B) - sin(A - B) (d) sin(A + B) + sin(A - B) (e) sin8T + sin2T (f) cos2x - cos4x (g) cos10° - 3 (h) 1 + cos20° 2 2 x-y x+y 2. (a) 2sin4T . cos2T (b) 2sin 2 . cos 2 (c) 2sin x+y . sin y-x (d) 2sin x+y . cos x-y 2 2 2 2 (e) 2 cos5° (f) -2sin15° . sin55° (g) 2sin55° . sin15° (h) 2cos4x . sinx (i) 2sin9x . sin2x (j) 2cos 7x . cos 3x 2 2 9.4 Conditional Trigonometric identities Identities which are true under given certain conditions are known as conditional identities. In this section, we deal with some trigonometric identities like A + B + C = Sc, then A + B = Sc - C, B + C = Sc - A, C + A = Sc - B We can write, (i) sin (A + B) = sin (Sc - C) = sinC sin(B + C) = sin (Sc - A) = sinA sin(C + A) = sin (Sc - B) = sinB (ii) cos (A + B) = cos(Sc - C) = - cosC cos (B +C) = cos (Sc - A) = - cosA cos(C + A) = cos (Sc - B) = - cosB (iii) tan(A + B) = tan (Sc - C) = - tanC tan (B + C) = tan(Sc - A) = - tanA tan(C + A) = tan (Sc - B) = - tanB (b) If A + B + C = Sc, then A + B + C = S2c. 2 2 2 A B C B C A C A B So, we write 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2 2 2 2 We write, ((i) (= sin sin A + B (( (= sin Sc - C = cos C (sin 2 2 (( 2 2 2 B C A A 2 + 2 Sc - 2 = cos 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233

vedanta Excel In Opt. Mathematics - Book 10 (sin C A (= sin Sc B B ((ii) 2 + 2 (= cos (2-2= cos 2 cos A + B (( Sc - C = sin C (cos 2 2 (((= cos 2 2 2 B C (( A A (cos 2 + 2 (= cos(( Sc - 2 = sin 2 (( 2 C A (( B B ((iii) tan2 + 2 (= tan(( Sc - 2 = sin 2 . 2 A B C C (tan 2 + 2 (= tan Sc - 2 = cot 2 , 2 B C A A (tan 2 + 2 (= tan Sc - 2 = cot 2 2 C A B B 2 + 2 Sc - 2 = cot 2 2 Worked Out Examples Example 1. If A + B + C = Sc, prove that Solution: (a) tanA + tanB + tanC = tanA . tanB . tanC  (b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1  2 2 2 (a) tanA + tanB + tanC = tanA. tanB.tanC  Example 2. Here, A + B + C = Sc. or, A + B = Sc - C ? tan (A + B) = tan (Sc - C) or, tanA + tanB = – tanC 1 – tanA.tanB or, tanA + tanB = – tanC + tanA.tanB.tanC  ? tanA + tanB + tanC = tanA. tanB.tanC proved. (b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1 2 2 2 Here, A + B + C = Sc or, A + B = Sc – C 2 2 2 2 tan(A2 B2 ) tan(S2c C  ? + = – 2 ) tan A + tan B C 2 2 2 (or, A B = cot 1 - tan 2 . tan 2 or, tan A + tan B = 1 - tanA2 . tan B 1 2 2 2 tan C 2 tanA2 C B C tanA2 B or, . tan 2 + tan 2 . tan 2 = 1 - . tan 2 .  ? tanA2 . tan B + tan B . tan C + tanC2 tanA2 = 1 Proved. 2 2 2 If A + B + C = 180°, then prove the following: (a) sin2A + sin2B + sin2C = 4sinA.sinB.sinC (b) cos2A + cos2B + cos2C = - 4cosA.cosB.cosC - 1 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Solution: (c) cos2A + cos2B - cos2C = 1 - 4sinA.sinB. cosC (a) Here, A + B + C = 180° sin (A + B) = sin (180° - C) = sinC and cos(A + B) = cos (180° - C) = - cosC ( (LHS = sin2A + sin2B + sin2C 2A - 2B 2A +2B 2 = 2sin 2 ((. cos + 2sinC.cosC. (( = 2sin(A + B). cos(A -B) + 2sinC.cosC (( = 2sinC . cos (A -B) + 2sinC.cosC = 2sinC[cos(A – B) + cosC] = 2sinC[cos(A-B)- cos (A + B)] = 2sinC. 2sinA.sinB = 4sinA - sinB. sinC = RHS. Proved. (b) Here, A + B + C = 180° ? cos(A + B) = cos(180° - C) = - cosC ( (LHS = cos2A + cos2B + cos2C 2A - 2B = 2cos 2A + 2B . cos 2 + 2cos2C - 1 2 = 2cos(A + B). cos(A -B) + 2cos2C - 1 = 2(-cosC). cos (A -B) + 2cos2C - 1 = -2cosC [cos(A-B) - cosC] - 1 =-2cosC. [cos(A - B) + cos(A + B)]- 1 = - 2cosC.[2cosA.cosB]- 1 = -4cosA . cosB . cosC - 1 = RHS. Proved. (c) Here, A + B + C = 180° cos(A + B) = cos(180° - C) = -cosC ( (LHS = cos2A + cos2B - cos2C = 2cos 2A+2B . cos 2A - 2B - 2cos2C + 1 2 2 = 2cos(A + B).cos(A - B) - 2cos2C + 1 = 2cos(A + B).cos(A -B) - 2cos2C + 1 = - 2cosC [cos (A - B) + cosC] + 1 = -2cosC[cos(A - B) - cos(A + B)] + 1 = - 2cosC. 2sinA.sinB + 1 = 1 - 4 sinA. sinB. cosC = RHS. Proved. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 235

vedanta Excel In Opt. Mathematics - Book 10 Example 3. If A + B + C = 180°, then prove the following: Solution: (a) sinA + sinB + sinC = 4cos A cs.ocinossAA22 B2.. .sciconossB2B2 C2.. ssiinnC2C2 236 (b) cosA + cosB + cosC = 1 + 2 (c) cosA + cosB - cosc = - 1 + 4 4 Here, A + B + C = 180° ( ( ( (Now, A + B (( C C 2 2 (( 2 2 sin = sin 90° - = cos and cos A + B = cos 90° - C = sinC2 2 2 2 (a) LHS = sinA + sinB + sinC ( (= (= A+2- B2sin+C2 s,incoCs 22scions C2A. +2coBs . cos( C A-B ( 2 ( 2(( [ ( ]= 2cosC2 cos A-B + sinC2 2 ( ( ]= 2cosC2 [cos A-B + cos A+B ( 2 2 = 2cosC2 2cosA2 B . . cos 2 = 4 cosA2 . cos B . cos C2 =RHS, Proved. 2 (b) LHS = cosA + cosB + cosC ( (= (= 2cos A+ B (. cos A-B +1 - 2sin2 C 2 (A-B 2 2sin2 2 C ( C 2sin 2 . cos (( 2 +1- 2 [ ( ]= 2sin C . cos A-B - sin C +1 2 2 2 [ ( ( ]= 2 sinC2 cos A-B - cos A+ B ( +1 2 2 C 2sinA2 B = 2sin 2 . . sin 2 + 1 = 1 + 4sinA2 . sin B . sin C = RHS. Proved. 2 2 (c) LHS = cosA + cosB - cosC ( (= 2cos A+ B (. cos A- B - cosC 2 ( 2 (= C (A- B C 2sin 2 .cos (( 2 - 1 + 2sin2 2 [ ( ]= 2 sinC2 cos A-B + sinC2 - 1 2 [ ( ( ]= 2 sinC2 cos A-B + cos A+B ( -1 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = 2sin C . 2cosA2 . cos B - 1 2 2 = - 1 + 4 sinC2 . cosA2 . cosB2 = RHS. Proved. Example 4. If A + B + C. = 180°, then prove the following: Solution: (a) sin2A + sin2B + sin2C = 2 + 2cosA.cosB.cosC (b) cos2A + cos2B + cos2C = 1 - 2cosA.cosB.cosC. (c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC We have, A + B + C = 180° or, A + B = 180° - C ? sin (A + B) = sin(180° - C) = sinC and cos(A +B) = cos(180° - C) = -cosC Now, (a) LHS = sin2A + sin2B + sin2C = 1 [2sin2A + 2sin2B] + 1 - cos2C 2 1 = 2 [1 - cos2A + 1 - cos2B] + 1 - cos2C 1 - 1 (cos2A + cos2B) - cos2C ( (= + 1 2 1 2A+2B 2A - 2B 2 2 2 = 2 - .2 cos ((, cos . - cos2C (( = 2 - cos(A + B).cos (A - B) - cos2C = 2 + cosC . cos(A - B) - cos2C = 2 + cosC [cos(A - B)- cosC] = 2 + cosC [cos(A - B) + cos (A + B)] = 2 + 2cosA. cosB. cosC = RHS. Proved. (b) LHS = cos2A + cos2B + cos2C = 1 [2cos2A + 2cos2B] + cos2C 2 1 = 2 [1 + cos2A + 1 + cos2B] + cos2C 1 [2 + (cos2A + cos2B)] + cos2C ( (=2 1 2A+2B 2A-2B =1+ 2 .2.cos 2 . cos 2 + cos2C = 1 + cos(A +B) . cos(A -B) + cos2C = 1 + (- cosC). cos (A -B) + cos2C = 1 - cosC [cos(A -B)- cosC] = 1 - cosC [cos(A -B) +cos(A +B)] =1 - cosC . 2cosA . cosB = 1 - 2cosA. cosB. cosC =RHS. Proved. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 237

vedanta Excel In Opt. Mathematics - Book 10 (c) LHS = sin2A - sin2B + sin2C = 1 [2sin2A - 2sin2B] + sin2C = 1 [1–cos2A–1+cos2B] + sin2C 2 2 1 2 (cos2B - cos2A) + sin2C ( (= 1 2B+2A 2A-2B 2 2 2 = . 2 sin (( . sin + sin2C ( = sin(A +B). sin(A -B) +sin2C (( = sinC. sin(A -B) + sin2C (( = sinC [sin (A - B) + sinC](( = sinC [sin (A -B) + sin (A + B)](( = sinC. 2sinA. cosB = 2 sinA cosB.sinC = RHS. Proved. Example 5. If A + B + C = Sc, then prove that Solution: (a) sin2 A + sin2 B - sin2 C = 1 - 2cos A . cos B sin C 238 2 2 2 2 2 2 A B C A B C (b) cos2 2 + cos2 2 - cos2 2 = 2cos 2 . cos 2 sin 2 Here, A + B + C = Sc or, A+B = Sc - C 2 2 ( (sin Sc C A+B = sin 2 - 2 = cos C 2 2 ( (cos Sc C C A+B = cos 2 - 2 = sin 2 2 Now, (a) LHS = sin2 A + sin2 B - sin2 C 2 2 2 ] ]=1 A B C 2 2 sin2 2 + 2sin2 2 - sin2 2 = 1 [1 - cosA + 1 - cosB] - sin2 C 2 2 1 C =1 - 2 [cosA + cosB] - sin2 2 ( (= 1 - 1 .2cos A+B . cos A-B - sin2 C 2 2 2 2 ] ( ]= C A-B C 1 - sin 2 cos 2 + sin2 2 ] ( ( ]= 1 - sin C cos A-B + cos A+B 2 2 2 C A B = 1 - sin 2 . 2cos 2 . cos 2 = 1 - 2cos A . cos B . sin C = RHS. Proved. 2 2 2 A B C 2 + 2 - cos2 2 ] ](b)LHS = cos2 cos2 = 1 2cos2 A + 2cos2 B - cos2 C 2 2 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = 1 [1 + cosA + 1 + cosB] - 1 + sin2 C 2 2 1 C = 1+ 2 [cosA + cosB] – 1 + sin2 2 ( (= 1 A+B A-B C 2 2 2 2 . 2cos ((. cos + sin2 (= ( sin C . cos A-B ( + sin2 C 2 2 2 (( ] ( ]= C sin 2 . cos A-B + sin C 2 2 ] ( ( ]=C sin 2 . cos A-B + cos A-B 2 2 C A B = sin 2 . 2cos 2 . cos 2 = 2cos A . cos B . sin C = RHS. Proved. 2 2 2 Example 6. (If A + B + C = Sc, prove thatABC Sc - A (cos Sc - B (. cos Sc - C Solution: 2 2 2 4 4 4 cos + cos + cos = 4cos (( ( Example 7. Solution: Here, A + B + C = Sc or, A + B = Sc - C B + C = Sc - A C + A = Sc - B ? cos(C + A) = cos(Sc – B) = –cosB Now, LHS = cos A + cos B + cos C 2 2 2 = 2cos A/2 + B/2 . cos A/2 – B/2 + cos C + cosS2c 2 2 2 ( ( ( (= 2 cos A+B A-B C + Sc C - Sc 4 4 4 4 (( . cos + 2cos . cos(( ( ( ( (= 2cos Sc - C A-B Sc + C C - Sc 4 4 4 4 . cos(( + 2cos . cos(( ( ] ( ( ]= 2cos Sc - C A-B Sc + C 4 4 4 (( ( cos + cos ( cos(–T) = cos T ( ( (= 2cos Sc - C A-B+Sc+ C A-B-Sc- C 4 8 8 ((( .2cos . cos ( ( (= 4cos Sc - C Sc+A+C -B A-(B + C) - Sc 4 8 8 (( cos . cos ( ( ] ( (= 4cos Sc - C Sc+Sc-B -B A-Sc + A - Sc 4 8 8 ((( cos . cos ( ( (= 4cos Sc - C Sc - B A - Sc 4 4 4 ( ((. cos . cos ( ( (= 4cos Sc - A Sc - B Sc - C 4 4 4 .cos( (( cos = RHS. Proved. If A + B + C = Sc, then prove that cos (B + C - A) + cos (C + A - B) + cos (A + B - C) = 1 + 4cosA. cosB.cosC. We have A + B +C = Sc or, A + B = Sc - C, B + C = Sc - A, C + A = Sc - B Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 239

vedanta Excel In Opt. Mathematics - Book 10 Now, LHS = cos (B + C - A) + cos (C + A - B) + cos (A + B - C) = cos (Sc - A - A) + cos (Sc - B - B) + cos (Sc - C - C) = cos (S - 2A) + cos (Sc - 2B) + cos(Sc - 2C) = - cos2A - cos2B - cos2C ( (= - [cos2A + cos2B] - (2cos2C - 1) 2A+2B 2A-2B 2 2 = - 2cos (((. cos - 2cos2C + 1 (( = 1 - 2cos (A + B).cos(A -B)- 2cos2C = 1 + 2cosC. cos(A -B) - 2cos2C = 1 + 2cosC [cos(A - B) - cosC] = 1 + 2cosC [cos(A - B) + cos (A +B)] = 1 + 2cosC.2cosA. cosB = 1 + 4 cosA. cosB. cosC = RHS. Proved. Example 8. If A + B + C = 180°, then prove that Solution: ( ( (sin(B+2C)+sin(C+2A)+sin(A+2B) = 4sin B-C .sin C-A .sin A-B 2 2 2 Here, A + B + C = 180° B + C = 180° - A C + A = 180° - B , A + B = 180° - C Now, LHS = sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = sin (B + C + C) +sin (C + A + A) + sin (A + B + B) = sin (Sc - A + C) + sin (Sc - B +A) + sin (Sc - C + B) = sin{Sc - (A - C)} + sin {Sc -(B - A)} + sin {Sc - (C - B)} = sin(A -C) +sin(B -A) + sin (C - B) ( (= 2sin A-C+B-A A-C-B+A 2 2 (( . cos - sin (B - C) ( ( ( (= 2sin B-C 2A - B - C B-C B-C 2 2 2 2 (( (. cos ( – 2sin . cos ( ( ] ( ( ]= 2sin B-C 2A - B - C B-C 2 2 2 (( ( cos – cos ( ] ( ( ]=2sin B-C 2A - B – C + B – C B - C – 2A+ B + C 2 4 4 (( 2sin sin ( ( (= 2sin B-C 2A - 2C 2B - 2A 2 4 4 (( (. sin . sin ( ( (= 4sin B-C A-C B-A ( 2 2 2 ((. sin . sin ( sin(–T) = –sin T ( ( (= 4 sin B-C C-A A-B 2 2 2 . sin((( . sin = RHS. Proved. 240 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 9. ( (If A + B + C = 180°, proved thatA A Solution: tan 2 + tan B+C (= sec 2 .sec B+C ( 2(( 2 Here, A + B + C = Sc, (( ( (sin B+C (= sin Sc - C = cos C 2 2 2 2 ( (and C cos B+C = cos Sc - 2 = sin C 2 2 2 (LHS A B+C = tan 2 + tan 2 = tan A + cotA2 = sinA/2 + cosA/2 2 cosA/2 sinA/2 sin2A/2 + cos2A/2 = sinA/2.cosA/2 = 1 = 1 . 1 sinA/2.cosA/2 sinA/2 cosA/2 (= cosecA2 . secA2 = sec A .sec B+C (= RHS. Proved. 2 2 Exercise 9.4 Very Short Questions 1. (a) Define conditional trigonometric identities with an example. (b) If A,B and C are angles of a triangle ABC, then show that. (i) sin(A + B) - sinC = 0 (ii) cos (B + C) + cosA = 0 (iii) tan(B + C) + tanA = 0 (c) If A + B + C = 180°, Show that : (sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = sin (A - C) + sin (B - A) + sin(C - B) C (d) If A, B, C are angles of a triangle, show that tan A+B ( = cot 2 2 Short Questions 2. If A + B + C = Sc, prove that (a) tanA + tanB + tanC - tanA tanB.tanC = 0 (b) cotA.cotB + cotB. cotC + cotC . cotA - 1 = 0 (c) tan A ..ctoatnB2B2. + tan B . tan C + tan C . tan A =1 (d) cot A2 2 2 2 2 2 C = A + B C cot 2 cot 2 cot 2 + cot 2 (e) tan2A + tan2B + tan2C = tan2A . tan2B . tan2C Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 241

vedanta Excel In Opt. Mathematics - Book 10 Long Questions 3. If A + B + C = Sc, prove the following : (a) sin2A + sin2B - sin2C = 4cosA.cosB.sinC (b) sin2A - sin2B + sin2C = 4cosA.sinB.cosC (c) cos2A + cos2B - cos2C = 1 - 4 sinA.sinB.cosC 4. If A + B + C = Sc, prove the following : (a) sinA - sinB + sinC = 4sin A .cos B . sin C 2 2 2 A B C (b) sinA + sinB - sinC = 4sin 2 .sin 2 . cos 2 (c) sinA - sinB - sinC = – 4cos A .sin B .sin C 2 2 2 A B C (d) - sinA + sinB + sinC = 4cos 2 .sin 2 .sin 2 5. If A + B + C = 180° then prove that the following : (a) cosA - cosB + cosC = 4cos A .sin B .cos C -1 2 2 2 A B C (b) -cosA + cosB + cosC = -1 + 4 sin 2 .cos 2 .cos 2 (c) cosA - cosB - cosC = 1 - 4sin A .cos B . cos C 2 2 2 6. If A + B + C = Sc, prove the following : (a) sin2A + sin2B - sin2C = 2sinA.sinB.cosC (b) sin2A - sin2B - sin2C = - 2cosA.sinB.sinC (c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC 7. If A + B + C = Sc, prove the following : (a) cos2A - cos2B + cos2C = 1 - 2sinA.cosB. sinC (b) cos2A - cos2B - cos2C = - 1 + 2cosA.sinB.sinC (c) cos2A + cos2B - cos2C = 1 – 2sinA.sinB.cosC 8. If A + B + C = Sc, prove the following : (a) sin2 A + sin2 B + sin2 C = 1 - 2sin A .sin B .sin C 2 2 2 2 2 2 A B C A B C (b) sin2 2 - sin2 2 + sin2 2 = 1 - 2cos 2 .sin 2 .cos 2 (c) sin2 A - sin2 B - sin2 C = 2sin A .cos B .cos C - 1 2 2 2 2 2 2 9. If A + B + C = Sc, prove the following : ( )(a) A B C A B C cos2 2 + cos2 2 + cos2 2 = 2 1 + sin 2 . sin 2 . sin 2 (b) cos2 A - cos2 B - cos2 C = - 2sin A . cos B .cos C 2 2 2 2 2 2 A B C A B C (c) cos2 2 - cos2 2 + cos2 2 = 2cos 2 . sin 2 .cos 2 242 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 10. If A + B + C = Sc, prove that (a) sin A + sin B + sin C = 1 + 4sin Sc- A . sin Sc- B . sin Sc- C . 2 2 2 4 4 4 A B C Sc-A Sc-B Sc-C (b) cos 2 + cos 2 + cos 2 = 4cos 4 . cos 4 . cos 4 (c) cosA + cosB + cosC = 1 + 4cos Sc-A cos Sc-B cos Sc-C 2 2 2 11. If A + B + C = Sc, prove that sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4sinA.sinB.sinC 12. If A + B + C = Sc, prove that ( ) ( ) ( )(a) cos(B+ 2C) + cos(C + 2A) + cos(A + 2B) = 1 - 4cos A -B .cos B-C .cos C -A 2 2 2 ( ) ( ) ( )(b) sin(B+ 2C) + sin(C + 2A) + sin(A + 2B) = 4sin A -B C -A 2 .sin B-C .sin 2 2 (c) sin3A + sin3B + sin3C = – cos 32A.cos32B.cos32C (d) cos4A + cos4B +cos4C = – 1 + 4cos2A . cos2B . cos23C 13. If D+ E + J = Sc , prove that 2 (a) sin2D + sin2E + sin2J = 1 - 2 sinD. sinE.sinJ (b) tanE.tanJ. + tanJ . tanD + tanD . tanE = 1 (c) cos(D - E - J) + cos (E - J - D) + cos(J - D - E) = 4cosD.cosE.cosJ 14. If A + B + C = Sc, prove that (a) sin2A + sin2B + sin2C = 8 sinA2 sinB2 sin C 4cosA2 .cosB2 .cosC2 2 (b) cosA + cosB + cosC = 2 sinB.sinC sinC.sinA sinA.sinB (c) sinA + sinB + sinC = 2tanA. tanB.tanC cosB.cosC cosC.cosA cosA.cosB (d) cosA.sinB.sinC + cosB.sinC.sinA + cosC.sinA.sinB = 1 + cosA.cosB.cosC. (e) sinA.cosB.cosC + sinB.cosC.cosA + sinC.cosA.cosB = sinA.sinB.sinC 15. If A + B + C = Sc, prove that tanA + tanB + tanC + tanB + tanC + tanA tanB tanC tanA tanA tanB tanC = cosA.secB.secC + cosB.secC.secA + cosC.secA.secB 16. If A + B + C = 2S, prove that sin(S – A).sin(S – B) + sinS.sin(S – C) = sinA.sinB Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 243

vedanta Excel In Opt. Mathematics - Book 10 9.5 Trigonometric Equations Let us consider the following equalities: (a) sin2T + cos2T = 1 (b) 4sin2T = 1 (c) sec2T = 1 + tan2T (d) tanT = 1 3 Discuss the following questions from above statements. (a) Which are identities ? (b) Which are equations ? (c) What are differences between identities and equations ? Equalities like sin2T + cos2T = 1, sec2T = 1 + tan2T are satisfied by all values of the angle T. So they are called identities. But 4sin2T = 1 and tanT = 1 etc. are satisfied by only some values of the angle T. So they 3 are called trigonometric equations. Definition: An equation containing the trigonometrical ratios of an unknown angle is called trigonometric equation. Example : sinT = 3 , 0° ≤ T ≤ 360° 2 For 0° ≤ T ≤ 360°, sinT = 3 2 or, sinT = sin60° or sin120° ? T = 60° or 120° To solve such trigonometric equations, we need some limitations for the value of variable. This limitation is called the range of the variables. We note that the values of sinT and cosT repeat after an interval Sc. If the equation involves a variable T, 0 ≤ T ≤ 2Sc, then the solutions are called principal solutions. A general solution is one which involves the integer 'n' and gives all solutions of trigonometric equation. But we study only about principal solutions. Example : Find the solution of cosT = 1 , 0° ≤ T ≤ 360° 2 We know that cosT = cos(360° - T) = cosT or cosT = 1 = cos30° 2 ? T = 30° or, cosT = 1 = cos(360° - 30°) = cos30° or cos330° 2 244 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Note : Let T be any angle. Then we note the following : i. The value of sinT lies between -1 and +1. Y i.e. -1 ≤ sinT ≤ 1, otherwise there is no solution. (180° – T),S (360° + T), ii. The values of cosT lies between -1 and (sin and cosec positive) (All positive) X' X +1. i.e. -1≤ cosT ≤ 1, otherwise there is TO C no solution. (tan and cot positive) ( cos and sec positive) iii. The value of tanT is infinitely positive or (180° + T) (360° – T) infinitely negative. it has no boundary of Y' its values. i.e. - ∞ < tanT < + ∞ iv. Find the quadrants where the sign of value of trigonometric ratio of angle falls by using the rule CAST. Some Basic Ideas i. If sinT = sinA, then T = A or 180° - A Example : Solve : sinT = 1 2 1 sinT = 2 = sin45° Since sine is positive in the first and the second quadrants, we write, sinT = sin45° or sin(180° - 45°) ? T= 45° or 135° ii. If sinT = - sinA, then T = 180° + A or, 360° - A Example : Solve sinT = - 1 2 sinT = - sin45° Since sine is negative in the third and the fourth quadrant we write, sinT = sin (180° + 45°) or sin(360° - 45°) ? T = 225° or 315° iii. If cosT = cosA, thenT= A or (360° - A) Example : Solve cosT = 3 2 ? cosT = cos30° Since cos is positive in the first and the fourth quadrant, we write, cosT = cos30° or cos(360° - 30°) ? T = 30° or, 330° iv. If cosT = -cosA, then T = 180° - A or, 180° + A Example : Solve : cosT = - 3 , cosT = - cos30° 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 245

vedanta Excel In Opt. Mathematics - Book 10 Since cos is negative in the second quadrant and third quadrant, we write, cosT = cos (180° - 30°) or cos (180° + 30°) ? T = 150° or 210° Working steps for solution of Trigonometric equations The following are some hints that will be helpful in solving a trigonometric equations : (i) Express all the trigonometric functions in terms of a single trigonometric function of same angle if possible. (ii) Transfer every terms of the equation to the left hand side. (iii) If equation is quadratic, factorize the left hand side and equate each factor to zero. (iv) If the equation is quadratic in certain trigonometric function use formulae for the solution (in case not factorizable). Note : In the process of solving trigonometric equations if there is involvement of squaring or cubing, which give rise to some additional equations and consequently some additional roots. All the solutions thus obtained may not satisfy the given equation. The values which do not satisfy are discarded: only the values which satisfy be given equations are accepted. Each is to be checked. Only accepted values are the solution of the given equation. Example : sinT - cosT = 1 0 ≤ T ≤ 360° Solution: We can write, sinT = 1 + cosT 246 Squaring on both sides, we get sin2T = 1 + 2cosT + cos2T or, cos2T + 2cosT + 1 - 1 + cos2T = 0 or, 2cos2T + 2cosT = 0 or, 2cosT (cosT + 1) = 0 Either, 2cosT = 0 ................ (i) cosT + 1 = 0 ........... (ii) From equation (i), cosT = 0 or, cosT = cos90°, cos270°, ? T = 90°, 270° or, cosT = -1 =cos180° ? T=180° Hence, T = 90°, 180°, 270° on checking, put T = 90°, in given equation sin90° - cos90° = 1 sin90° - cos90° = 1 or, 1 = 1 (It is true.) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Putting T = 180°, sin180° - cos180° = 1 or, 0 + 1 = 1 ? 1 = 1 (true) Putting T = 270. sin270° - cos270° = 1 or, -1 -0 = 1 or, -1 = 1 (false) Hence, 270° is rejected. ? T = 90°, 180° Some special cases for 0° ≤ T ≤ 360° (i) If sinT = 0, then T= 0°, 180°, 360° (ii) If cosT= 0, thenT = 90°, 270° (iii) If tanT = 0, thenT = 0, 180°, 360° (iv) If sinT = 1, then T = 90° (v) if sinT = -1, then T = 270° (vi) If cosT = -1, then T = 180° (vii) If tanT = ∞, then T = 90°, 270° (viii)90° × n ± , when n = odd number (i.e. 1, 3, 5, .....) sin œ cos, tan œ cot, sec œ cosec when n = even number (i.e. 2, 4, 6, .............) sin œ sin, cos œ cos, tan œ tan Worked Out Examples Example 1. Solve the equation: (0° ≤ T ≤ 180°) Solution: (a) sinT = 3 , (b) tanT = 3 (c) cosT = 1 (a) Here, 2 = 2 3 sinT 2 since, 0 ≤ T ≤ 180° sinT = 3 = sin60° or sin(180° - 60°) 2 ? T = 60° Ÿ 120° (b) Here, tanT = 3 since 0° ≤ T ≤ 180° Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247

vedanta Excel In Opt. Mathematics - Book 10 tanT = 3 = tan60° ? T = 60° (c) Here, cosT = 1 2 since, 0° ≤ T ≤ 180° cosT = 1 = cos45° 2 ? T = 45° Example 2. Solve the equation: 0° ≤ T ≤ 360°. Solutions: (a) 3 tanT = 1, (b) cosec2T = 2 (c) 3tan2T - 1 = 0   (a) Here, 3 tanT = 1 or, tanT = 1 3 Since tanT is positive, T lies in the first or the third quadrant. ? tanT = tan30° or, tan (180° + 30°) ? T= 30° or, 210° (b) cosec2T = 2 or, 1 =2 or, sin2T = 1 sin2T 2 1 ? sinT = ± 2 Taking positive sign, we get sinT = 1 2 Since, sinT is positive, T lies in the first or second quadrant, sinT = sin45° or, sin(180° - 45°) ? T = 45° or, 135° Taking negative sign, we get sinT = - 1 2 Since sinT is negative it lies in the third or fourth quadrant.  ? sinT = sin(180° + 45°) or, sin(360° - 45)  ? T= 225° or, 315° Hence, the required values of T are 45°, 135°, 225° or 315° (c) Here, 3 tan2T - 1 = 0 ( )or, 1 2 tan2T = 3 or, tanT = ± 1 3 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Taking positive sign, tanT = 1 3 Since tanT is positive, Tlies in the first or third quadrant ? tanT= tan30° or, tan (180° + 30°) ? T = 30° or 210° Taking negative sign, we get tanT = - 1 3 Since tanT is negative, T lies in the second or fourth quadrant. ? tanT = tan(180° - 30°) or, tan (360° - 30°) ? T = 150° or, 330° ? T = 30°, 150°, 210°, 330°. Example 3. Solve : tan 9T = cotT, 0° ≤ T ≤ 90° Solution: Here, tan 9T = cotT or, tan9T = tan(90° - T), [cotT = tan(90° - T)] tan(270° - T), tan(7×90° - T), tan(5×90° - T), tan(9×90° - T) ? 9T = 90° - T, 270° - T, 450° - T, 630° - T, 810° - T or, 10T = 90°, 270°, 450°, 630°, 810° ? T = 9°, 27°, 45°, 63°, 81° Example 4. 4sin2T + 3cosT = - 3, 0° ≤ T ≤ 360° Solution: Here, 4sin2T + 3cosT = - 3 or, 4 - 4cos2T + 3cosT + 3 = 0 or, - 4cos2T + 3cosT + 7 = 0 or, 4cos2T – 3cosT – 7 = 0 or, 4cos2T - 7cosT + 4cosT - 7 = 0 cosT (4cosT - 7) + 1 (4cosT - 7) = 0 or, (4cosT - 7) (cosT + 1) = 0 Either, 4cosT - 7 = 0 ........... (i) cosT + 1 = 0 ............... (ii) From equation (i) cosT = 7 >1 4 since, - 1 ≤ cosT ≤ 1, it has no solution. Hence, cosT = 7 is rejected. 4 From equation (i), we get, cosT = - 1 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249

vedanta Excel In Opt. Mathematics - Book 10 or, cosT = cos180° ? T = 180° Hence, the required solution is T = 180°. Example 5. 2cosT = 3 cotT, 0° ≤ T ≤ 360° Solution: Here, 2cosT = 3 cotT Example 6. Solution: or, 2cosT = 3 cosT (Note: Do not cancel cosT) sinT or, 2cosT.sinT - 3 cosT = 0 or, cosT (2sinT - 3 ) = 0 Either, cosT = 0 .............(i) or, 2sinT - 3 = 0 .............. (ii) From equation (i), cosT = 0 or, cosT = cos90°, cos270° ? T= 90°, 270° From equation (ii), sinT = 3 2 or, sinT = sin60° or, sin (180° - 60°) ? T = 60° or, 120° Hence, the required solutions are x = 60°, 90°, 120°, 270° Solve : 3 sinT - cosT = 1, 0° ≤ T ≤ 360° Here, 3 sinT - cosT = 1............ (i) coefficient of sinT = 3 coefficient of cosT = - 1 ( 3)2 + (-1)2 = 3 + 1 = 2 Dividing equation (i) by '2' on both sides, we get, 3 sinT - 1 cosT = 1 2 2 2 1 or, cos30° . sinT - sin30° cosT = 2 or, sin(T - 30°) = sin30° or, sin (180° - 30°) ? T - 30° = 30° or, 150° or, T = 30° + 30° or, 150° + 30° ? T = 60°, 180° are required solutions. Alternative Method Here, 3 sinT = 1 + cosT Squaring on both sides, we get, 3sin2T = 1 + 2cosT + cos2T or, 3(1 - cos2T) = 1 + 2cosT + cos2T 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 or, 2cosT + cos2T + 1 = 3 - 3cos2T or, 4cos2T + 2cosT - 2 = 0 or, 2cos2T + cosT - 1 = 0 or, 2cos2T + 2cosT - cosT - 1 = 0 or, 2cosT (cosT + 1) - 1 (cosT + 1) = 0 or, (2cosT - 1) (cosT + 1) = 0 Either, 2cosT - 1 = 0 ........ (i) cosT + 1 = 0 .......(i) From equation (i), cosT = 1 , 2 or, cosT = cos60° or, cos(360° - 60°) T = 180° ? T = 60° or 300° T = 300° is rejected From equation (ii) cosT = -1, or, cosT = cos180° ? ? T = 60°, 180°, 300° On checking, For, T = 60° 3 sinT - cosT = 1 or, 3 . sin60° - cos60° = 1 or, 3 . -2312- 1 = 1 or, 3 2 2 =1 or, 1 = 1 (True) For T q or, 3 sin180° - cos180° = 1 or, 3 . 0 - (-1) = 1 ? 1 = 1 (True) For T = 300° or, 3 sin300° - cos300° = 1 ( )or, 3 - 3 - 1 =1 2 2 3 1 or, - 2 - 2 =1 or, -2 = 1 (False) ? ? T = 60°, 180° are required values. Example 7. Solve cos3T + cosT = cos2T, 0° ≤ T ≤ 180° Solution: Here, cos3T + cosT = cos2T Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 251

vedanta Excel In Opt. Mathematics - Book 10 ( ) ( )or, 2cos 3T + T . cos 3T - T = cos2T 2 2 or, 2cos2T.cosT - cos2T = 0 or, cos2T(2cosT - 1) = 0 Either, cos2T = 0 ........ (i) 2cosT - 1 = 0 ........ (ii) From equation (i), cos2T = 0 or, cos2T = cos90°, cos270° ? 2T = 90°, 270° ? T = 45°, 135° From equation (ii), 2cosT - 1 = 0 or, cosT = 1 2 or, cosT = cos60° ? T = 60°, 300° ? T 45°, 60°, 135° are the required values. Example 8. Solve 3 + 1 = 4 , (0° ≤ T≤ 90°) Solution: sin2T cos2T Example 9. Here, 3 + 1 = 4 Solution: sin2T cos2T or, 3 cos2T + sin2T = 4cos2T.sin2T Dividing both sides by 2, we get, 3 cos2T + 1 sin2T = 2sin2T.cos2T 2 2 or, sin60°. cos2T + cos60°. sin2T = sin4T or, sin(2T + 60°) = sin4T or, sin4T = sin(2T + 60°) or, 4T = 2T + 60° or, 2T = 60° ? T= 30° Solve : tanT + tan2T + tanT . tan2T = 1, 0° ≤ T ≤ 360° Here, tanT + tan2T + tanT.tan2T = 1 or, tanT + tan2T = 1 - tanT. tan2T or, tanT + tan2T = 1 1 - tanT . tan2T or, tan 3T= tan45°, tan (180° + 45°), tan(360° + 45°), tan(540° +45°), tan(720° + 45°), tan(900° + 45°) 252 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10  ? 3T = 45°, 225°, 405°, 585°, 765°, 945° ? T = 15°, 75°, 135°, 195°, 255°, 315° Example 10. Solve : tan2T - (1 + 3 ) tanT + 3 = 0, (0° ≤ T ≤ 360°) Solution: Here, tan2T - (1 + 3 ) tanT + 3 = 0 or, tan2T - tanT - 3 tanT + 3 = 0 or, tanT (tanT - 1) - 3 (tanT - 1) = 0 or, (tanT - 1) (tanT - 3) = 0 Either, tanT - 1 = 0 ....... (i) tanT - 3 = 0 ........ (ii) From equation (i), tanT = 1 or, tanT= tan45°, tan225° From equation (ii), tanT = 3 or, tanT = tan60°, tan240° ? T = 60°, 240°  ? T = 45°, 60°, 225°, 240° are the required values. Example 11. If tanT + tanE = 2 and cosv.cosE = 1 solve for values of D and E 2 Solution: Here, tanD + tanE = 2 or, sinD + sinE =2 cosE cosD or, sinD.cosE + cosD.sinE = 2cosD. cosE or, sin(D + E) = 2cosD. cosE ....... (i) But cosD.cosE = 1 ......... (ii) 2 1 ? sin(D + E) = 2 .2 or, sin(D + E) = 1 = sin90° ? D + E = 90° and E = 90° - D From (ii) 2cosD.cos(90° - D) = 1 or, 2cosD. sinD = 1 or, sin2D = sin90° or, 2D = 90° ? D = 45° Also, D + E = 90° Ÿ E = 90° - D = 90° - 45° = 45° ? D = 45°, E = 45° Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 253

vedanta Excel In Opt. Mathematics - Book 10 Exercise 9.5 Very Short Questions 1. Find the value of T in the following equation, 0° ≤ T ≤ 180°. (a) 2sinT = 3 (b) 3 tanT = 1 (c) cosT = 1 (d) cotT = 3 (e) secT = 2 2 (f) 2cosT + 1 = 0 (g) 3 tanT - 1 = 0 2. Find the value of T, 0° ≤ T ≤ 90° (a) tanT = cotT (b) cosT = sinT (c) secT = cosecT (d) cot2T = tanT (e) sin4T = cos2T (f) cos2T = sin3T 3. Solve the following questions : 0° ≤ T ≤ 180° (a) sin2T = sinT (b) tanT = sinT (c) 3cotT - tanT = 2 (d) 2 +2=0 (e) tanT + cotT = 2 (f) 1 + cos2T = cosT cosT Short Questions 4. Solve, 0° ≤ T ≤ 360° (b) 4sin2T = 3 (c) sec2T = 4 (a) 3tan2T = 1 (e) 3sec2T - 4 = 0 (d) sec2T = 2tan2T Long Questions 5. Solve for T, 0° ≤ T ≤ 180° (a) sin4T = cosT - cos7T (b) sinT + sin2T + sin3T = 0 (d) cos2T + cos4T = cos3T (c) sinT - sin2T = 0 (f) 3cotT - tanT = 2 (e) secT . tanT = 2 6. Solve : 0° ≤ T ≤ 360° (b) 2sin2T - 3sinT + 1 = 0 (a) 2cos2T - 3sinT  (c) 2cos2T - 5cosT + 2 = 0 (d) 2sin2T + sinT - 1 = 0 (e) 4cos2T + 4sinT = 5 (f) 3 - 2sin2T = 3cosT ( )(h) cot2T + (g) tan2T+ (1 - 3 ) tanT = 3 3 1 cotT + 1 = 0 7. Solve : 0° ≤ x ≤ 360° 3 (a) 3 cosx + sinx = 3 (b) sinx + cosx = 2 (d) 3 cosx = 3 - sinx (c) sinx + cosx = 1, (f) cosx - 3 sinx = 1 (e) 3 sinx + cosx = 2 (g) 2 secx + tanx = 1 8. Solve for x, 0° ≤ x ≤ 180° (a) 3 + 1 =0 (b) 3 + 1 = 4 sin2x cos2x sin2x cos2x 254 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 9. Solve : 0° ≤ x ≤ 360° (a) 2 tan3x.cos2x + 1 = tan3x + 2cos2x (b) sin2x . tanx + 1 = sin2x + tanx 10. Find the value of T from. tanT - 3cotT = 2tan3T, 0° ≤ T ≤ 360° 11. Solve: 0° ≤ x ≤ 90° tan(x + 100°) = tanx.tan(x + 50°).tan(x – 50°) 1. (a) 60°, 120° (b) 30° (c) 60° (d) 30° (e) 45° (f) 120° (g) 30° 2. (a) 45° (b) 45° (c) 45° (d) 30° (e) 15°, 75° (f) 18°, 90° 3. (a) 0°, 60°, 180° (b) 0°, 180° (c) 45°, 108.43° (d) 135° (e) 45° (f) 60°, 90° 4. (a) 30°, 150°, 210°, 330° (b) 60°, 120°, 240°, 300° (c) 60°, 120°, 240°, 300° (d) 45°, 135°, 225°, 315° (e) 30°, 150°, 210°, 330° 5. (a) 0°, 10°, 45°, 50°, 90°, 130°, 135°, 170° (b) 0°, 90°, 120°, 180° (c) 0°, 60°, 180° (d) 30°, 60°, 90°, 150°, 180° (e) 45°, 135° (f) 45°, 108.43° 6. (a) 30°, 150° (b) 0°, 30°, 90° (c) 60°, 300° (d) 30°, 150°, 270° (e) 30°, 150° (f) 0°, 60°, 300°, 360° (g) 60°, 135°, 240°, 315° (h) 120°, 150°, 300°, 330° 7. (a) 0°, 60°, 360° (b) 45° (c) 0°, 90°, 360° (d) 0°, 60°, 360° (e) 60° (f) 0°, 240° (g) 315° 8. (a) 60°, 150° (b) 20°, 30°, 50°, 80°, 120°, 140°, 170° 9. (a) 15°, 30°, 75°, 150°, 210°, 225°, 315°, 330° (b) 45°, 225° 10. 45°, 60°, 120°, 135°, 225°, 240°, 300°, 315° 11. 30°, 55° 9.6 Height and distance Trigonometric knowledge can be used to calculate heights of all objects or distance of inaccessible objects. A Introduction : Let AB be a pillar. An observer is at point C which is at a distance BC from 12cm the foot of the pillar. The observer observes the top of the pillar through CA. The line CA is called sight line. AB is the height of the pillar. BC is the B 60° distance between the foot of the pillar and the observer. Let AC = 12m and C ‘BCA = 60°. Now discuss the following questions. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 255

vedanta Excel In Opt. Mathematics - Book 10 (i) What is the height of the pillar ? (ii) Find the distance between the foot of the pillar and the observer. Let us define some basic terms used in this topic : Object A Angle of elevation : When an object is at a higher level than that of the Sight line observer, the observer has to have a view of the object Angle of elevation or the observer has to look up. The angle which the sight line makes with the horizontal the observers Observe's eye O T eye is known as the angle of elevation. In the figure ‘AOB = T is the angle of elevation. O Horizontal Line through B T observe's eye A Angle of Depression : When the object is at a lower level than that of the observer Angle of Depression Sight line then he has to look below in order to see the object. The C B angle which the sight line through the observer's eye is known as the angle of depression. In the above figure ‘AOB = T is the angle of depression. OA is parallel to CB, ‘AOB = ‘CBO = T, being the corresponding angles. How is the angle of elevation or depression measured? Special instruments like theodolite, sextant, clinometer, hypsometer etc are used to measure the angle of depression. Step 1 : Draw the appropriate figure which represents the given condition. Step 2 : Identify the given sides and angles in the figure. Step 3 : Choose the appropriate trigonometric ratio to find the unknown sides or angles. Step 4 : Carry out the necessary calculations with appropriate units. Worked Out Examples Example 1. Find the values of x and y from the given figures: (a) A (b) S 45° P xx D 30° 60° B Q 40m R 20m Cy Solution: (a) In the given figure from right angle ∆ABC, tan60° = x y or, 3 y = x ? x = 3 y ............ (i) 256 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Again, from right angled ∆ABD, tan30° = 20 x y + or, 1 = x y 3 20 + or, 20 + y = 3 x .......... (ii) Putting the value of x in equation (i), we get 20 + y = 3. 3y or, 20 + y = 3y or, 2y = 20 ? y= 20 ? y = 10m 2 Putting the value of y in equation (i) 20 + 10 = 3 x or, x= 30 = 10 3 ? x = 10 3m 3 (b) In the given figure, SP//QR, corresponding angles. ‘SPQ = ‘PQR, Now, from right angled triangle PQR, tan45° = x or, 1 = x 40 40 ? x = 40m Example 2. A man observes the angle of elevation of the top of a tower to be 30°. On Solution: walking 200 metre nearer, the elevation is found to be 60°. Find the height of the tower. Also, find the distance between the first point from the foot of the tower. Let CD be the height of the tower, A and B be the first and the second point respectively. In the figures AB = 200m, suppose BC = ym and CD = xm. From right angled triangle BCD, tan60° = CD BC or, 3 = x y D or, x = 3 y .......... (i) Again, from right angled triangle ACD. x tan30° = CD 30° 60° AC 200m By A C x or, 1 = 200 + y 3 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 257

vedanta Excel In Opt. Mathematics - Book 10 or, 200 + y = 3 x ........... (ii) Putting the value of x in equation (ii) 200 + y = 3. 3 y or, 200 + y = 3y or, 200 = 2y ? y = 100m Putting the value of y in equation (i) x = 3. 100 = 1.732 × 100 = 173.2 m ? The height of the tower is 173.2m and distance between the tower and first point is (200 + 100) = 300m Example 3. From the roof of a house, an observer finds that the angles of depression of Solution: two places in the same side are 60° and 45°. If the distance between these places is 40m, find the distance of the places from the foot of the house. Let PQ be the height of the house and R and S he two points on the same side of the house. Then PT//QS, ‘TPR = ‘PRQ = 60°, ‘TPS = ‘PSR = 45°, RS = 40m. Let QR = y m and PQ = x m. P 60°45° T Now, from right angled triangle PRQ tan60° = PQ xm QR x or, 3 = y Q 60° 45° S y R 40m or, x = 3 y ............... (i) Again, from right angled triangle PQS tan45° = PQ QS x or, 1 = y + 40 or, x = y + 40 ........ (i) Putting the value of x in equation (ii) from (i), we get 3 y = y + 40 or, y ( 3 - 1) = 40 or, y = 5434.06-14m= 40 - 1 = 40 ? y = 1.732 0.732 Putting the value of y in equation (i) x = 3 × 54.64 = 94.64m ? The height of the house (x) = 94.64m Distance of the first point from the house (y) = 54.64m 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Distance of the second point from the house (x) = y + 40 = 54.64 + 40 = 94.64m. Example 4. Two girls are on the opposite sides of a tower which is 200m high. They Solution: observed that the angles of elevation of the top of the tower are 30° and 45° respectively. Find the distance between them. Let A and B be the positions of the two girls on opposite sides of the tower DC. DC = 200m. The angles of elevations are ‘DAC = 30°, ‘DBC = 45° Let AC = xm and BC = ym From right angled ∆DAC, D tan30° = DC 200m AC 1 200 or, x =3 = x A 30° 45° B or, 200 3 .......(i) xm C ym Again, from right angled ∆ACB, tan45° = CD CB 200 or, 1= y ? y = 200m = AB = AC + CB Distance between the girls = x + y = 200 3 m + 200m = 546.41m Example 5. The angle of depression of the top and the foot of a lamp post observed from Solution: the roof of a 60m high house and found to be 30° and 60°. Find the height of the lamp post. Let MN and PQ be the height of the house and lamp post respectively. Then, MN = 60m, ‘RMP = ‘MPS = 30°, ‘RMQ = ‘MQN = 60° From right angled, ‘MQN, R 30° M 60° MN tan60° = QN or, 3 = 60 P 30° S QN 60 60 ? QN = 3 = 3 3 = 20 3 m. Q 60° N PS = QN = 20 3 (opposite sides of a rectangle) Again, from right angled triangle MPS, tan30° = MS PS 1 MS or, 3 = 20 3 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 259

vedanta Excel In Opt. Mathematics - Book 10 MS = 20 3 = 20m 3 Hence, the height of the lamp post is PQ = SN = MN - MS = 60 - 20 =40m. Example 6. A boy standing between two pillars of equal height observes the angle of Solution: elevation of the top of pillars to be 30°. Approaching 15m towards any one of the pillar, the angle of elevation is 45°. Find the height of the pillars and the distance between them. Let PQ and MN be two equal pillars of equal height, QN the distance between the pillars. O is mid point of QN. MN = PQ. ‘MON = 30°, ‘POQ = 30°, RO = 15m. From right angled triangle PQR, tan45° = PQ QR or, QR = PQ Again, from right angled triangle PQO, tan30° = PQ P M QO N 1 PQ or, 3 = QR + 15 1 PQ Q 45° 30° 30° 3 PQ + 15 R 15m O or, = or, 3 PQ = PQ + 15 or, ( 3 - 1 ) PQ = 15 or, PQ = 15 = 15 × 3 +1 3-1 3-1 3 +1 15 ( 3 +1) = 3-1 = 15 (1.732 + 1) = 7.5 × 2.732 2 = 20.49m. Again, from the right angled triangle MON, tan30° = MN or, 1 = 20.49 ON 3 ON ? ON = 35.49m Distance between two pillars is 2 × ON = 2 × 35.49 = 70.98m. Hence, the height of the pillar = PQ = MN = 20.49m 260 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 7. A flagstaff stands upon the top of a building. Find the length of the flagstaff Solution: and building observed from a point at a distance of 20 m from the building are respectively 60° and 45°. Let PQ be the flagstaff and QR be the building respectively. P Let S be a point 20m away from the building, SR = 20m Now, from right angled triangle SRQ. Q tan45° = QR 60° SR 45° S R 20m or, 1= QR 20 ? QR = 20m Again, from right angled triangle PRS. tan60° = PR SR or, 3 = PQ + QR SR or, 3= PQ + 20 20 or, 20 3 = PQ + 20 ? PQ = 20 ( 3 - 1) = 20 (1.732 - 1) = 20 × 0.732 = 14.64m. Hence, the height of the flagstaff is 14.64m. Example 8. PQ and MN are two towers standing on the same horizontal plane. From Solution: point M the angle of depression of Q is 30° and from N the angle of elevation of P is 60°. If the shorter tower is MN = 100m, find the height of the tower PQ. In given two towers are PQ and MN, MN<PQ. MN = 100m, ‘RMQ = ‘MQN = 30° ‘PNQ = 60°. P From right angled triangle MQN, tan30° = MN QN or, 1 = 100 R 30° M 3 QN Q 30° 60° N 100m ? QN = 100 3 m. 261 RMNQ is a rectangle, Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 ? RM = QN = 100 3 m. From right angled triangle PQN, tan60° = PQ QN PQ or, 3= 100 3 ? PQ = 300m ? The height of the longer tower PQ is 300m. Example 9. From a light house the angles of depression of two ships on the opposite Solution: sides of the light house were observed to be 30° and 45°. If the height of the light house is 200m and the line joining the two ships passes through the foot of the light house, find the distance between the ships. Let AD be the height of the light house. ‘EAB = ‘ABD = 45°, ‘FAC = ‘ACD = 30° From the right angled triangle ABD, tan45° = AD BD or, BD = AD A ? BD = 100m E F 100m Again, from right angled triangle ADC, tan30° = AD B 45° D 30° C DC or, 1 = 100 3 DC or, DC = 173.2m. Hence, the distance between the ships is given by BC = BD + DC = 100m + 173.2m = 273.2m Example 10. A ladder of 9 m long reaches a point 9 m below the top of a vertical flagstaff. From the foot of the ladder the angle of elevation of the flagstaff is 60°. Find the height of the flagstaff. Solution: Let MN be the height of the flagstaff and PS be a ladder. Then ‘MPN = 60°, MS = 9m, PS = 9m M Let SN = x, PN = y . From right angled triangle MPN, by using Pythagoras theorem, 9m PS2 = PN2 + SN2 P 60°y S or, 81 = y2 + x2 x ? x = 81 - y2 ......... (i) N 262 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Again, from right angled triangle MPN, tan60° = MN PN or, 3= 9+x y or, 3 y = 9 + x ............ (ii) By solving equation (i) and (ii), we get 3 y – 9 = 81 – y2 squaring on both sides, we get, ( 3 y – 9)2 = ( 81 – y2)2 or, 3y2 – 18 3y + 81 = 81 – y2 or, 3y2 + y2 = 18 3y or, 4y2 = 18 3y or, 4y2 = 18 3 y 18 3 or, y = 4 since y ≠ 0, y = 93 = 7.8 2 and x = 81 – 60.75 = 4.5m ? MN = 9 + 4.5 = 13.5m Alternative Method In above diagram, ∆PSM is an isosceles triangle.  ? ‘PMS = ‘MPS = 30° ‘SPN = 60° - 30° = 30° From, right angled triangle ∆SPN sinSPN = SN PS x or, sin60° = 9 or, 1 = x 2 9  ? x = 4.5 m Hence, the height of the flagstaff is (9 + 4.5) = 13.5m Example 11. A vertical column is divided into ratio 1:9 from the bottom by a point on it. If these two parts make an equal angle 20m from the bottom of the any column, find the height of the column. Solution: Let MN be the height of the vertical column and P be the point on it which divides MN in the ratio of 1:9 from the bottom. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 263

vedanta Excel In Opt. Mathematics - Book 10 Let NP = x and PM = 9x. Let each segment of the column makes an angle T at Q. ‘PQN = T‘MQP = T, QN = 20m M From right angled triangle PQN 9x tanT = PN = x P QN 20 x Again, from right angled triangle MQN, N tan2T = MN Q T QN T 20m or, 2tanT = 10x or, 1- -ta2nx02T 20 1 = x x2 2 1 - 400 or, x × 400 = x 10 400 - x2 2 or, 400 - x2 = 80 or, x2 = 320 Since height is positive, taking positive square root only, x = 8 5 m. The height of the column is 10x = 10 × 8 5 = 80 5 m. Example 12. The angle of elevation of a bird from a point 80m above the lake is 30°. From the same point, the angle of depression to its image on the same lake is found to 60°. Find the height at which the bird is at the moment. Solution: Let B be the position of the bird and its image is B' in the lake. Then a point A is 80m above the surface of water, XY the level of water of the lake. Let BC = x, CB' = y, AD = 80m. ‘CAB = 30°, ‘B'AC = 60° From right angle triangle ABC, B x tan30° = BC C 30° CA 1 x 60° A or, 3 = CA XE 80m DY or, CA = 3 x ........(i) B' Again, from right angled triangle ACB' or, tan60° =3y.xCCBA'[by using (i)] 3= or, y = 3x ............ (i) But y = CE + EB' ( BE = B'E) 264 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 B'E = 80 + BC = 80 + x ? y = 80 + 80 + x ....... (iii) From (ii) and (iii), we get, 160 + x = 3x or, 2x = 160 ? x = 80m ? The position of the bird from the level of lake is BE = CE + CB = 80 + 80 = 160m Example 13. The shadow of a pole on the ground level increases in length by x metres when the sun's attitude changes from 45° to 30°. Calculate values of x given that height of the pole is 50m. Solution: Let MN be the height of the pole, MN = 50m. Let NP = y and NQ = y + x be the length of shadows when sun's altitudes are 45° and 30° respectively. Increased in length of shadow is PQ = x. From, right angled triangle MPN, M 50m tan45° = MN PN 50 or, 1 = y or, y = 50m Q 30° 45° N Again, from right angled triangle MQN. xPy tan30° = MN QN or, 1 = 50 3 x+y or, 50 3 = x + 50 or, x = 50 ( 3 - 1) or, x = 50 × (1.732 - 1) = 50 × 0.732 = 36.6m ? x = 36.6m Exercise 9.6 Very Short Questions 1. Define the following terms with figures: (a) Angle of elevation (b) Angle of depression 2. Find the value of x in the following diagrams: PM 30° N 20m (a) (b) 60m x Q 60° R P xO Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 265

vedanta Excel In Opt. Mathematics - Book 10 Short Questions 3. Find the values of x and y in the given diagrams: P (a) (b) P y 45° S 30° 45° Q 200m 20m R x Q 60° R x Sy (c) P (d) P 30° 60° U Q 9m 9m 20m 30° R 60° S T x 60° S x y y R Q Long Questions 4. (a) The angle of elevation of the top of a house from a point on the ground was observed to be 60°. On walking 60m away from that point, it was found to be 30°. If the house and points are in the same line of the same plane, find the height of the house. (b) The angle of elevation of the top of a tower from a point was observed to be 45°. On walking 30m away from that point, it was found to be 30°. Find the height of the house. 5. (a) From a helicopter flying vertically above a straight road, the angles of depressions of two consecutive kilometer stone on the same side are found to be 45° and 60°. Find the height of the helocopter. (b) From the top of a tower of 100m, the angle of depression of two places due to east of it are respectively 45° and 60°. Find the distance between the two places. (c) From the top of a tower of 200 m, the angle of depression of two boats which are in a straight line on the same side of the tower are found to be 30° and 45°. Find the distance between the boats. 6. (a) From the top of 21 metre high cliff, the angles of depression of the top and bottom of a towers are observed to be 45° and 60° respectively. Find the height of the tower. (b) From the top of a cliff the measure of the angle of depression of the top and bottom of a building are found to be 30° and 45°. If the height of the cliff is 100 metres. Find the height of the building. 7. (a) Two poles stand on either side of a road. At the point midway between the two posts, the angles of inclination(elevation) of their tops are 30° and 60°. Find the length of the shorter posts if the longer post is of 15m long. 266 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (b) Two lamp posts are of equal height. A girl standing mid way between them observes the angle of elevation of either posts to be 45°. After walking 15m towards one of them, she observes its elevation to be 60°. Find the heights of the posts and the distance between them. (c) The posts are 120m apart, and the height of one is double that of other. From the middle point of the line joining the feet, an observer found the angle of elevation of their tops to be complementary. Find the height of the longer post. 8. (a) A flagstaff stands upon the top of a building. The angles of elevation of the top of the building and flagstaff are observed to be 45° and 60° respectively. If the distance between the observer and the building is 10m, find the length of the flagstaff. (b) A flagstaff stands on the top of a tower. The angles subtended by the tower and the flagstaff at a point 100 metres away from the foot of the tower are found to be 45° and 15°. Find the length of the flagstaff. (c) A flagstaff is placed at one corner of a rectangular garden 40m long and 30m wide. If the angle of elevation of the top of the flagstaff from the opposite corner is 30°. Find the height of the flagstaff. 9. (a) From the top and bottom of a tower, the angle of depression of the top of the house and angle of elevation of the house are found to be 60° and 30° respectively. If the height of the building is 20m. find the height of the tower. (b) From the roof of a house 30m height, the angle of elevation of the top of a tower is 45° and the angle of depression of its foot is 30°. Find the height of the tower. (c) A crow is sitting on the top of a tree which is in front of a house. The angle of elevation and angle of depression of the crow from the bottom and top of the house are 60° and 30° respectively. If the height of the tree is 21m. Find the height if the house. 10. (a) A ladder of 18m reaches a point of 18m below the top of vertical flagstaff. From the foot of the ladder the angle elevation of the flagstaff is 60°. Find the height of the flagstaff. (b) A ladder 15m long reaches a point 15 below the top of a vertical flagstaff. From the foot of the ladder, the angle of elevation of the top of flagstaff is 60°. Find the height of the flagstaff. 11. (a) A pole is divided by a point in the ratio 1:9 from bottom to top. If the two parts of the pole subtend equal angles at 20m away from the foot of the pole, find the height of the pole. (b) AB is a vertical pole with its foot B on a level of ground. A point C on AB divides such that AC : CB = 3:2. If the parts AC and CB subtend equal angles at a point on the ground which is at a distance of 20m from the foot of the pole, find the height of the pole. 12. (a) A man 1.75m stands at a distance of 8.5m from a lamp post and it is observed that his shadow is 3.5m long. Find the height of the lamp post. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 267

vedanta Excel In Opt. Mathematics - Book 10 (b) A tree of 12m height is broken due to wind such a way that its top touches the ground and makes an angle of 60° with the ground. At what height from the bottom is the tree broken by the wind. 13. The angle of elevation of an aeroplane from a point in the ground is 45°. After 15 seconds of flight the angle if elevation changes to 30°. If the aeroplane is flying horizontally at a height of 4000m in the same direction, find the speed of aeroplane. 14. The angle of elevation of the top of a tower is 45° from a point 10m above the water level of a lake. The angle of depression of its image in the lake from the same point is 60°. Find the height of the tower above the water level. 15. A rope dancer was walking on a loose rope tied to the tops of two posts each 8m high. When the dancer was 2.4m above the ground, it was found that the stretched pieces of the rope made angles of 30° and 60° with the horizontal line parallel to the ground. Find the length of the rope. Project Work 16. Select buildings around your school. Find the angle of elevation by using sextants or theodolite. Find the height of the buildings taking the distance between the two places in meter. 17. Write a short note on \" Height and Distance\" with its application. 2. (a) 30 3 m (b) 20 3 m 3. (a) x = y = 27.32 m (b) x = 200 m, y = 115.47 m (c) x = 7.79 m, y = 4.5 m (d) x = 20 3, y = 40 m 4. (a) 51.96 m (b) 40.98 m 5.(a) 2365.98 m (b) 42.26 m (c) 146.4 m 6.(a) 8.87 m (b) 42.27 m 7. (a) 5 m (b) 35.49 m, 70.98 m (c) 84.85 m 8. (a) 7.32 m (b) 73.2 m (c) 28.86 m 9. (a) 80 m (b) 81.96 m (c) 28 m 10. (a) 27 m (b) 22.5 m 11.(a) 178.88 m (b) 22.36 m 12. (a) 6 m (b) 5.57 m 13. 195.2 m/s 14. 37.32 m 15. 17.67 m 268 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 10Vectors 10.0 Review Discuss the following questions: a. Classify the following quantities to vectors or scalars with your reasons. (i) Temperature (ii) Pressure (iii) Force (iv) Length (v) Area (vi) Acceleration (vii) Displacement (viii) Time b. Give an example of each of the following vectors: (i) Column vector (ii) Row vector (iii) unit vector (iv) Equal vectors (v) Negative vectors (vi) position vectors (vii) Unequal vectors (viii) Like and unlike vectors. c. What role do vectors play in our daily life ? Give some examples. o o oo oo d. Let a = (2, -3) and b = (1, 2), find a + b and a - b . Express them in the oo form of x i + y j . 10.1 Scalar (or Dot) Product of Two Vectors oo Let OA = (2, 3), OB = (-3, 2) be two vectors. Plot them in a graph paper. Then, (a) Find the product of x-components and y-components Y of the vectors. (b) Add the products. (c) What is the result ? BA X oo X' O (d) Are OA and OB perpendicular to each other? Here, 2 × (-3) + 3 × 2 = - 6 + 6 = 0 oo OA is perpendicular to OB . There are two types of product of two vectors: Y' (a) Scalar product or dot product. (b) Vector product or cross product. We discuss only about scalar product or dot product of two vectors. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 269

vedanta Excel In Opt. Mathematics - Book 10 Scalar Product of Two Vectors The scalar product of two vectors oa and ob is defined as the product of the magnitude of two vectors multiplied by the cosine of the angle T between their directions. Thus, oa · ob = |oa | |ob | cosT = ab cosT where, |oa |= a and |ob |= b. AO = oa and OB = ob Now, draw perpendicular BM from B to OA. Now, oa · ob = |oa | |ob |cosT = a b cos T = (OA) (OB) cos T = OA (OB cos T) = (OA) (OM) = (magnitude of oa ) (component of ob in the direction of oa ) So, it is clear that the scalar product of two vectors is equivalent to the product of the magnitude of one vector with the component of the other vector in the direction of this vector. If we write oa · ob , the rotation of oa towards ob is anti-clockwise and the angle T is taken to be positive. ? oa · ob = ab cosT If we write ob ;oa , the rotation of ob towards oa is clockwise and the angle T is taken to be negative. ? ob · oa = b a cos(–T) = ba cosT Hence, oa · ob = ob · oa Note : Let oa = x1oi + y1oj and ob = x2oi + y2oj be two vectors in terms of the components. Then we know that oi and oj are unit vectors along X-axis and Y-axis respectively. oi . oj = |oi ||oj |cos 90° = 0, oi . oi = |oi | |oi | cos0° = 1 Now, oa . ob = (x1 oi + y1) . (x2oi + y2oj ) = x1oi .(x2oi + y2oj ) + y1 oj . (x2oi + y2 oj ) = x1x2 oi . oi + x1y2 oi . oj + y1 x2 oj . oi + y1y2 oj . oj = x1x2 + y1y2 ? oa . ob = x1x2 + y1y2 Hence, the scalar product of two vectors is equal to the sum of the product of their corresponding components. 270 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Two Important Results of Scalar Product of Two Vectors. (a) If oa and ob are perpendicular to each other, i.e., T = 90° B A Then, oa . ob = |oa | |ob | cosT = |oa | |ob | cos90° = 0 o Hence, oa . ob = 0 b o a Conversely, of oa . ob = 0, then T = 90° oo (b) If a and b are parallel to each other, i.e., T = 0° or T = 180° o o |oa | |ob | cos0° = 1 a. b= and o o = |oa | |ob | cos180° = ab.(-1) = - ab a. b oo oo Conversely, if a . b = ± ab, T = 0° or 180°, a and b are parallel to each other. Some Important Results a. If |oa | = a and |ob | = b, oo a . b = abcosT. oo (i) |oa |= 0 (ii) |ob | = 0 or o o b. If a . b = 0, a A b oo c. If T = 0°, the value of a . b = ab is maximum. oo d. If T = 180°, the value of a . b = - ab is minimum. oo oo e. a . b = b . a . oo o o oo f. a . (k b ) = (k a ) . b = k( a . b ), where k z 0. k R. g. o o = |oa | |oa | cos0 = a.a. =a2 a. a Also, (i) o o = |oi | |oi | cos0° = 1.1.1 = 1 i. i (ii) o o = |oj | |oj | cos0° = 1.1.1 = 1 j. j (iii) o o = |oi | |oj | cos90° = 1.1.0 = 0 i. j o . o = |oj | |oi | cos90° = 1.1.0 = 0 (iv) j i Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 271

vedanta Excel In Opt. Mathematics - Book 10 Angle between Two Vectors Let us consider two points A(a1, a2) and B (b1, b2) in the plane. Then, position vector of A = OA = oa = a1 a2 = ob = b1 Position vector of B = OB b2 Magnitudes of oa and ob are | OA | = OA = a = |oa | | OB | = OB = b = |ob | Let ‘XOA = E, ‘XOB = D and ‘AOB = T. Then, ‘AOB = D – E = T. Draw perpendiculars AM and BN from A and B to the x-axis. Then, OM = a1, MA = a2, ON = b1 and NB = b2. From the right-angled triangle OMA, cosE = OM = aa1 ? a1 = a cos E OA ? a2 = a sinE aa2 sinE = MA = OA Similarly from the right-angled triangle ONB, b1 = b cos D and b2 = b sin D Now, a1b1 + a2b2 = a cos E b cosD + a sin E b sin D = ab cos(D – E) = |oa | |ob | cos T ................(i) But, by the definition of scalar product of two vectors, oa and ob oa · ob = | oa | |ob | cos T .............(ii) Now, from (i) and (ii) oa · ob = a1b1 + a2b2. This result leads us to define scalar product of two vectors in another way. Let oa = a1 and oa = b1 be two vectors. Then the scalar product of oa and ob is denoted a2 b2 by oa · ob and is defined by oa · ob = a1 b1 a2 · b2 = a1b1 + a2b2. Again from (i), |oa | |ob | cos T = a1b1 +a2b2 or, cos T = a1b1 + a2b2 or oa . ob |oa | |ob | |oa | |ob | oa ob . This result gives us angle between two vectors and cosT = a1b1 + a2b2 or oa . ob |oa | |ob | |oa | |ob | 272 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Worked out Examples oo oo Example 1. (a) If a = (4,5) and b = (2, 3), find the scalar product of a and b . Solution: (b) If |oa |= 4 and o = 3, T = 45°. find o o . Example 2. b a. b Solution: oo Example 3. (a) Here, a = (4, 5), b = (2, 3) Solution: Since (a1, a2). (b1,b2) = a1b1 + a2b2 oo a . b = (4, 5).(2,3) = 4.2 + 5.3 = 8+15= 23 oo (b) Here, | a | = 4, | b | = 3, T = 45° oo o o 1 = 6 2 a . b = | a | | b | cosT = 4.3 cos45° = 12. 2 o oo ooo If a = 2 i + j and b = i - 2 j , then, find the following: oo oo (i) a . b (ii) angle between a and b o o oo o o Here, a = 2 i + j , b = i - 2 j oo o o o o (i) a . b = (2 i + j ). ( i - 2 j ) = 2.1 + 1.(-2) = 2 - 2 = 0 oo oo Since a . b = 0, a and b are perpendicular to each other. o (ii) Here,| a | = 22 + 12 = 4 + 1 = 5 o | b | = 12+(-2)2 = 1 + 4 = 5 oo Let T be the angle between a and b , we get, cosT = oo = 0 =0 a. b 55 | oa || ob | ? T = 90° ooo o oo o (a) If a = 3 i + 4 j and b = -8 i + 6 j , show that a is perpendicular oo o oo to b , (b) If a = (2, 4) and b = (4, 8), show that a and b are parallel to each other. o o oo oo (a) Here, a = 3 i + 4 j , b = - 8 i + 6 j oo oo oo a . b = (3 i + 4 j ) . (-8 i + 6 j ) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 273

vedanta Excel In Opt. Mathematics - Book 10 = 3.(-8) + 4.6 = - 24 + 24 = 0 oo o o Since a . b = 0, a and b are perpendicular to each other. ( ) ( )(b) o 2 o 4 Here, a = 4 , b= 8 ( ) ( )o o 2 . 4 = 2.4 + 4.8 = 8 + 32 = 40 4 8 a. b = o | a | = 22 + 42 = 4 + 16 = 20 = 2 5 o | b | = 42 +82 = 16+64 = 80 = 4 5 oo Let T be the angle between a and b Now, cosT oo =2 40 5 = 1 = cos0° = a. b 5 ×4 oo | a || b | ? T = 0° oo Hence, a and b are parallel to each other. oo oo Example 4. (a) Find the value of m if two vectors m i + 7 j and 4 i + 8 j are Solution: perpendicular to each other. ( ) ( )o (b) If OA = 2 o -3 and ‘AOB = 90°, find the value of m. m and OB = 4 (a) Here, let o = o + o o oo a mi 7 j, b =4 i+8 j oo oo Since a and b are at right angle, a . b = 0 oo oo oo i.e. a . b = (m i + 7 j ) . (4 i + 8 j ) or, 0 = 4m + 56 or, 4m = - 56 ? m = -14 ( ) ( )(b) o o oo 2 , OB = b= -3 , ‘AOB = T = 90° Let OA = a = m 4 Then, by definition of dot product, ( ) ( )o o 2 . -3 = 2(-3) + m.4 = - 6 + 4m m 4 a. b= oo Since, T = 90° or, a . b = 0 or, - 6 + 4m = 0 or, 4m = 6 ? m= 3 2 274 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur