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vedanta Excel In Opt. Mathematics - Book 10 or, 4x - 12y + 16 = 0 ? x - 3y + 4 = 0 Taking negative sign, we get, 4x = (5y - 3) - 7y + 13 or, 4x + 2y - 10 = 0 ? 2x + y - 5 = 0 Therefore, the required separate equations are x - 3y + 4 = 0 and 2x + y - 5 = 0. Example 3. Find the equation of pair of lines represented by the equation Solution: x2 - 2xycosecD+ y2 = 0 Given equation is x2 - 2xy cosecD + y2 = 0 or, x2 - 2xy cosecD + y2 (cosec2D -cot2D) = 0 or, x2- 2xy cosecD + y2cosec2D - y2cot2D =0 or, (x - ycosecD)2 - (y cotD) 2 = 0. or, (x - y cosecD + y cotD) (x - ycosecD - y cotD) = 0 or, {x - y (cosecD - cotD)} {x-y(cosecD + cotD)} = 0 Hence, the required pair of lines are x - y (cosecD - cotD) = 0 and x - y (cosecD + cotD) = 0 Example 4. Find the angle between the pair of the lines represented by: Solution: (a) x2 + 7xy + 3y2 = 0 (b) x2 + xy - 6y2 + 7x + 31y- 18 = 0 (a) Given equation is 2x2 + 7xy + 3y2 = 0 ......... (i) Comparing it with ax2 + 2hxy + by2 = 0, we get, a = 3, h= 7 and b =3 2 Let T be the angle between the lines represented by equation (i). Then, tanT = ± 2 h2 - ab = ± 2 49 - 2.3 =± 49 -24 a+b 4 2.5 ? tanT = ± 1 2 +3 Taking positive sign, tanT = 1 ? tanT = tan45° ? T = 45° Taking negative sign, tanT = - 1 or, tanT = tan135° ? T = 135° Therefore, the required angles between the lines represented by equation (i) are 45° and 135° Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175

vedanta Excel In Opt. Mathematics - Book 10 (b) Here, x2 + xy - 6y2 + 7x + 31y - 18 = 0 Taking the homogeneous terms only, we get x2 + xy - 6y2 = 0 ........... (i) Comparing it with ax2 + 2hxy + by2 = 0, we get a = 1, h = 1 , b = -6. 2 Let T be the angle between the lines represented by equation (i), we get =±2 h2 - ab 1 + 6 1+24 =±2 4 4 tanT a+b =2 =±2 =±1 1-6 -5 1 2 .5 Taking positive sign, -5 tanT = 1 = tan45° ? T = 45° Taking negative sign, tanT = - 1 = tan 135° ? T = 135° Therefore, the required angles are 45° ans 135°. Example 5. Find the angle between the pair of lines represented by Solution: x2 - 2xy cosecD+ y2 = 0 Given equation is x2 - 2xy cosecD + y2 = 0 Example 6. Comparing it to ax2 + 2hxy + by2 = 0, we get Solution: a = 1, h = -cosecD, b- 1 Let T be the angle between the lines represented by (i), tanT = ±2 h2 - ab = ± 2 cosec2D = ±cotD a+b 1+1 = cot (± D) = tan(90° ± D) ? T = 90 ± D Show that the lines represented by 3x2 + 8xy - 3y2 = 0 are perpendicular to each other. Equation is 3x2 + 8xy - 3y2 = 0 ........... (i) comparing it with ax2 + 2hxy + by2 = 0, we get a = 3, h = 4, b = -3 Now, a + b = 3 - 3 = 0 Since a + b = 0, the lines represented by equation (i) are perpendicular to each other. 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Alternatively Here, 3x2 + 8xy - 3y2 = 0 or, 3x2 + 9xy - xy - 3y2 = 0 or, 3x(x + 3y) - y(x + 3y) = 0 or, (x + 3y) (3x - y) = 0 Separate equation of lines are : x + 3y = 0 ............(i) 3x - y = 0 ............. (ii) slope of line (i), m1 = - 1 3 slope of line (ii), m2 = .3 m1.m2 = 1 × 3 = - 1 3 ? m1.m2 = -1, the lines (i) and (ii) are at right angle each other. Example 7. Find the value of k when the pair of lines represented by (k+2) x2+8xy+ 4y2=0 are coincident. Solution: Given equation is (k + 2) x2 + 8xy + 4y2 = 0 comparing it to ax2 + 2hxy + by2 = 0, we get, a = k + 2, h = 4, b = 4 condition for coincidence h2 = ab or, 16 = (k + 2). 4 or, 16 = k + 2 4 or, 4 = k + 2 ? k=2 Example 8. Find the single equation passing through the point (1, 1) and parallel to the Solution: lines represented by the equation x2 - 5xy + 4y2 = 0 Given equation is x2 - 5xy + 4y2 = 0 or, x2 - 4xy - xy + 4y2 = 0 or, x(x - 4y) - y (x - 4y) = 0 or, (x - 4y) (x - y) = 0 Separate equations of lines are x - 4y = 0 ........... (i) and x - y = 0 .............. (ii) Equation of any line parallel to (i) is x - 4y + k1 = 0 ......... (iii) Equation of any line parallel to (ii) is x - y + k2 = 0 ......... (iv) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177

vedanta Excel In Opt. Mathematics - Book 10 Above both lines pass through the point (1, 1) 1 - 4 + k1 = 0 or, k1 = 3 and 1 - 1 + k2 = 0 or, k2 = 0 putting the values of k1 and k2 in equations (iii) and (iv), we get, x - 4y + 3 = 0 x-y =0 required single equation is given by, (x - 4y + 3) (x - y) = 0 or, x(x - 4y + 3) - y (x - 4y + 3) = 0 or, x2 - 4xy + 3x - 3x - xy + 4y2 - 3y = 0 ? x2 - 5xy + 4y2 + 3x - 3y = 0, which is the required single equation. Example 9. Find the single equation of straight line passing through the origin and Solution: perpendicular to the lines represented by x2 + 3xy + 2y2 = 0 Given equation is x2 + 3xy + 2y2 = 0 or, x2 + 2xy + xy + 2y2 = 0 or, x(x+2y) + y(x + 2y) = 0 or, (x + 2y) (x + y) = 0 Separate equations of lines are x + 2y = 0 and x + y = 0 Equations of the lines perpendicular to x + 2y = 0 and x + y = 0 and passing through the origin are respectively 2x - y = 0 and x - y = 0 Required single equation is (2x - y) (x - y) = 0 or, 2x2 - 3xy + y2 = 0 ? 2x2 - 3xy + y2 = 0 is the required equation. Example 10. Prove that two straight lines represented by the equation (x2 + y2)sin2D = (xcosE - ysinE)2 makes an angle is 2D. Solution: The given equation of two straight line is (x2 + y2)sin2D = (xcosE - ysinE)2 or, x2sin2D + y2sin2D = x2cos2E - 2xy sinE.cosE + y2sin2E or, x2sin2D - x2cos2E + y2sin2D - y2sin2E + 2xy sinE.cosE = 0 or, x2(sin2D - cos2E) + 2xy sinE.cosE + y2(sin2D - sin2E) = 0 By comparing the equation to ax2 + 2hxy + by2 = 0, we get a = sin2D - cos2E, 2h = sinE.cosE i.e. h = sinE.cosE 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 and b = sin2D - sin2E we know that tanT = ± 2 h2 - ab = ± 2 (sinE.cosE)2 - (sin2D - cos2E) (sin2D - sin2E) a+b sin2D - cos2E + sin2D - sin2E = ± 2sinD (1 - sin2D) 2sin2D - (sin2E + cos2E) 2sinD.cosD sin2D = ± 2sin2D - 1 = ± cos2D = ± tan2D or, tanT = tan2D ? T = 2D Therefore, the angle two lines represented by given equation is 2D. Proved. Exercise 8.2 Very Short Questions 1. (a) Write the general equation of second degree in x and y. (b) Define a homogeneous equation. (c) What is the angle between the pair of lines represented by ax2 + 2hxy + by2 ? (d) Write the conditions of coincident and perpendicularity of two lines represented by equation ax2 + 2hxy + by2 = 0 2. Find the single equation represented by the pair of straight lines: (a) x - y = 0 and x + y = 0 (b) x + 3y = 0 and x - 3y = 0 (c) 4x + 3y = 0 and x + y = 0 (d) 3x + 4y = 0 and 2x - 3y = 0 3. Find the separate equations of straight lines represented by the following equations: (a) x2 - y2 = 0 (b) 9x2 - 25y2 = 0 (c) x2 - y2 + x - y = 0 (d) x2 + y2 - 2xy + 2x - 2y = 0 4. Show that each pair of lines represented by equations represent a pair perpendicular lines: (a) 3x2 + 8xy - 3y2 = 0 (b) 5x2 + 24xy - 5y2 = 0 (c) 6x2 - 5xy - 6y2 = 0 5. Show that each pair of lines represented by equations are coincidence: (a) 16x2 + 16xy + 4y2 = 0 (b) x2 - 6xy + 9y2 = 0 (c) 4x2 - 12xy + 9y2 = 0 (d) 9x2 - 6xy + y2 = 0 Short Questions 6. Determine a single equation representing the following pair of lines. (a) x + y + 3 = 0 and x - 2y + 3 = 0 (b) 2x + y - 3 = 0 and 2x + y + 3 = 0 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179

vedanta Excel In Opt. Mathematics - Book 10 7. Determine the lines represented by each of the following equations: (a) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0 (b) x2 + 2xy + y2 - 2x - 2y - 15 = 0 (c) 2x2 + 3xy + y2 + 5x + 2y - 3 = 0 (d) 2x2 + xy - 3y2 + 10y - 8 = 0 8. Prove that two lines represented by the following equations are perpendicular to each other: (a) x2 - y2 + x - y = 0 (b) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0 9. Find the value of O when the lines represented by each of the following equations are coincident: (a) 4x2 + 2Oxy + 9y2 = 0 (b) (O + 2) x2 + 3xy + 4y2 = 0 10. Find the value of O when the following equations represents a pair of line perpendicular to each other: (a) Ox2 + xy - 3y2 = 0 (b) (O - 25) x2 + 5xy = 0 (c) (3O + 4) x2 - 48xy - O2y2 = 0 11. Find the angle between the following pair of lines: (a) x2 + 9xy + 14y2 = 0 (b) 2x2 + 7xy + 3y2 = 0 (c) x2 - 7xy + y2 = 0 (d) 9x2 - 13xy - 9y2 + 2x - 3y + 7 = 0 (e) x2 + 6xy + 9y2 + 4x + 12y - 5 = 0 12. Find the angle between the following pair of lines: (a) x2 - 2xy cotD - y2 = 0 (b) x2 + 2xy cosecD + y2 = 0 (c) y2 + 2xy cotD - x2 = 0 (d) x2 - 2xy cot2D - y2 = 0 13. Find the two separate equation of two lines represented by the following equations. (a) x2 + 2xy cotD - y2 = 0 (b) x2 - 2xy secD + y2 = 0 (c) x2 + 2xy cosecD + y2 = 0 (d) x2 - 2xy cot2D – y2 = 0 Long Questions 14. (a) Find the equation of two lines represented by 2x2 + 7xy + 3y2 = 0. Find the point of intersection of the lines. Also, find the angle between them. (b) Find the separate equations of two lines represented by x2 - 5xy + 4y2 = 0. Find the angle between the lines and the points of intersection of the lines. (c) Find the separate equation of two lines represented by the equation x2 - 2xy cosecD + y2 = 0. Also find the angle between them. 15. (a) Find the pair of lines parallel to the lines x2 - 3xy + 2y2 = 0 and passing through the origin. (b) Find the pair of lines parallel to the lines 2x2 - 7xy + 5y2 = 0 and passing through the point (1, 2). 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 16. (a) Find the pair of lines perpendicular to the lines x2 - 5xy + 4y2 = 0 and passing through the origin. (b) Find the equation of two straight lines which pass through the point (2,3) and perpendicular to the lines x2 - 6xy + 8y2 = 0 17. (a) Find the two separate equations when lines represented by kx2 + 8xy - 3y2 = 0 are perpendicular to each other. (b) Find the two separate equations when the lines represented by 6x2 + 5xy - ky2 = 0 are perpendicular to each other. 18. (a) Show that the pair of lines 3x2 - 2xy - y2 = 0 are parallel to the lines 3x2 - 2xy - y2 - 5x + y + 2 = 0 (b) Show that the pair of lines, 4x2 - 9y2 = 0 and 9x2 - 4y2 = 0 are perpendicular to each other. Project Work 19. Write short notes with examples on the following (a) linear equations (b) quadratic equations (c) homogeneous equations 2. (a) x2 - y2 = 0 (b) x2 - 9y2 = 0 (c) 4x2 + 7xy + 3y2 = 0 (d) 6x2 - xy - 12y2 = 0 3.(a) x + y = 0, x - y = 0 (b) 3x + 5y = 0, 3x - 5y = 0 (c) (x + y + 1) = 0, x - y = 0 (d) x - y = 0, x - y + 2 = 0 6.(a) x2 - xy - 2y2 + 6x - 3y + 9 = 0 (b) 4x2 + 4xy + y2 - 9 = 0 7.(a) x + 3y + 5 = 0, x + 3y - 1 = 0 (b) x + y - 5 = 0, x + y + 3 = 0 (c) 2x + y = 1, x + y + 3 = 0 (d) x - y + 2 = 0, 2x + 3y = 4 9.(a) ± 6 (b) - 23 16 10. (a) 3 (b) 25 (c) -1, 4 11. (a) T = tan–1(±13) (b) 45°, 135° (c) tan-1 ± 35 (d) 90° (e) 0° 12. (a) 90° (b) 90° ± D 2 (d) 90° (c) 90° 13. (a) x + y (cosecD + cotD) = 0, x - y (cosecD - cotD) = 0 (b) x + y(secD + tanD) = 0, x + (secD - tanD) = 0 (c) x + y(cotD + cosecD) = 0, x - y (cotD - cosecD) = 0 (d) x + y (cosec2D - cot2D) = 0, x - y (cosec2D + cot2D) = 0 14. (a) x + 3y = 0, 2x + y = 0, (0, 0), 45°, 135° (b) x - 4y = 0, x - y = 0, tan-1 ± 3 , (0, 0) 5 (c) x - y(cosecD + cotD) = 0, x + y(cotD - cosecD) = 0, 90° ± D 15. (a) x - 2y = 0, x - y = 0 (b) x - y + 1 = 0, 2x - 5y + 8 = 0 16. (a) 4x + y = 0, x + y = 0 (b) 4x + y = 11, 2x + y = 7 17. (a) 3x - y = 0, x + 3y = 0 (b) (3x – 2y) = 0, 2x + 3y = 0 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181

vedanta Excel In Opt. Mathematics - Book 10 8.3 Conic Section Conic sections are special types of locus which can be obtained geometrically as the intersection of a plane and a right circular cone. The following are basic types of conic sections. (a) Parabola (b) Ellipse (c) Circle (d) Hyperbola Conic sections are also called conics. They play vital roles in the field of mathematics, engineering, aeronautics etc. In present days, conics are very important in the study of basic level mathematics. The path of missile in the form of projectile and path of cables of suspension bridges are parabolic in nature. In present age of science and technology, parabolic metal surfaces are used in television dishes, telescopes, headlights, etc. The planets have elliptical paths while moving around the sun. A smooth elliptical surface can be used to reflect sound waves and light rays from one focus to another. Definition : A conic section is the intersection of a plane and a right circular cone. Before defining different types of conic section, let us define a right circular cone. Conic means \"cone\" and section means \"to cut\" or \"to divide\". 8.3 Cone A cone has a closed curved surface as its base Upper A axis AG and the lateral surface is generated by a line Nappie G T segment that rotates around the perimeter of Lower Semi vertical V the closed curved surface keeping one end Nappie fixed as the vertex. If the base is circle then D angle the cone is a circular cone. Generator From above figure we have the following V Vertex terms : Semi-vertical angle The angle Dis called semi-vertical angle. If we rotate the line VG about the line AV so that the angle D does not change. The surface generated is double napped right circular hollow cone. Vertex The point V is called the vertex. Axis The fixed line AV is called the axis of the cone. Generator The rotating line VG is called the generator of the cone. It is also slant height of the cone. Nappies The vertex divides the cone into two parts called nappies. 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 8.5 Conic section as a plane section of a right circular cone On intersecting a right circular cone by a plane in different positions, different sections so obtained are called conic sections or conics. The shape of conic section depends on the position of the intersecting plane with respect to cone and by the angle made by it with the axis of the cone. (a) Circle V Circle O The section of a right circular cone cut by a plane Plane parallel to its base represents a circle. In other words, when a cutting plane is at right angles to axis of the cone, then the section is called circle. (b) Parabola V Plane The section of a right circular cone by a plane parallel D to a generator of the cone represents a parabola. In T other words, when a cutting plane is inclined to the axis such that it is parallel to the generator of the cone, Parabola then the section so obtained is called a parabola. If T is a angle made by the plane with the axis of cone, O then D= T. (c) Ellipse D T Section of a right circular cone T cut by a plane not parallel to any generator and not parallel or not D perpendicular to the axis of the cone is called ellipse. In other words, 183 when a cutting plane is inclined at some angle to the axis of the cone, then the section so formed is called a ellipse. In this case D< T< 90°, where D is the semi-vertical angle and Tis the angle made by the plan ze with the axis of the cone. (d) Hyperbola When the double cone is cut with a plane parallel to the axis, we get hyperbola. In other words, when the cutting plane is inclined to the axis such that it also cuts the other part of Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 the double cone, the section is called hyperbola with two branches. In this case T< D, where Tis the angle made by the plane with the axis of the cone. Exercise 8.3 1. (a) Name conic section formed by a plane with right circular cone parallel to its base? (b) Name conic section formed by a plane with right circular cone parallel to its generator. (c) Name conic section formed by plane with right circular cone inclined at some angle to axis (but not 90°) 2. Name the following geometrical shapes: (a) (b) (c) (d) 3. Name the curves which is intersection of plane and cone in the following figures: (a) (b) V V (c) V (d) 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Project Work 4. Cutting a potato or cucumber, folding a paper how a parabola or an ellipse or a hyperbola can be formed. Discuss in the classes. 5. Take a potato or radish or carrot, make a base of right circular cone. Draw an axis of the cone on a paper. Then cut the slices of potato or radish or carrot perpendicular to the base, parallel to the base at different angles to axis. Identity the slices so formed parabola or ellipse or hyperbola or circle. 1. (a) circle (b) parabola (c) ellipse (d) hyperbola 2. (a) circle (b) ellipse (c) Parabola (d) hyperbola 3. (a) parabola (b) circle (d) ellipse 8.6 Circle Introduction : Let us draw a circle with centre at the Y origin and radius 5 units as shown in the graph. Discuss the answers of the following C (–3, 4) questions : (a) Take two points A(5,0) and B(-5, 0) on the O A (5, 0) X circumference find the mid-point of AB. Y' (b) Take point C(-3, 4) on the circumference, X' B (–5, 0) join OC. Find the length of OA, OB and OC, Are they equal ? Is there any common name for them ? (c) Is AB2 = AC2 + BC2 ? (d) Find the slopes of AC and BC, is there any relations of slopes of AC and BC? (e) What is equation of locus of points on above circle ? Discuss the answers of above questions and draw conclusions from the discussion. Definition of Circle The locus of a moving point which moves in such a way that its distance from a fixed point is always constant is said to be a circle. The fixed point is called centre of the circle and the constant distance is called the radius of the circle. In the above figure in graph, O(0,0) is the centre, OA, OB or OC are radii of the circle. Then, we write OA = OB = OC = r, radius of the circle. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185

vedanta Excel In Opt. Mathematics - Book 10 Equations of Circles Y (a) Equation of a circle with centre at the P(x, y) origin (standard form) r O Let us draw a circle with centre at the X' X origin. Let us take P(x, y) on the circle. Then OP = r, is the radius of the circle. Then by using distance formula. OP = r Squaring on both sides, we get OP2 = r2 Y' or, (x - 0)2 + (y - 0)2 = 0 ? x2 + y2 = r2 which is the required equation. Example 1. Find the equation of a circle with centre at the origin and radius 6 units. Solution: Here, radius of the circle, r = 6 units centre = 0(0,0) Equation of required circle is, x2 + y2 = r2 or, x2 + y2 = 62 ? x2 + y2 = 36 (b) Equation of a circle with centre at other than origin (Central Form) Let C(h, k) be the centre of circle and radius CP = r. Let Y P(x, y) be any point on the circle. Then, by using distance formula, P(x, y) We get C(h, k) CP = r X' X or, CP2 = r2 [squaring on both sides, we get] O or, (x - h)2 + (y - k)2 = r2 which is the equation of circle called central form. Y' Example 2. Find the equation of the circle whose centre is at (4, 5) and radius 6 units. Solution: Here, centre (h, k) = (4, 5) Radius (r) = 6 units Equation of required circle is (x - h)2 + (y - k)2 = r2 or, (x - 4)2 + (y - 5)2 = 62 or, x2 - 8x + 16 + y2 - 10y + 25 = 36 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 or, x2 + y2 - 8x - 10y + 5 = 0 The required equation of circle is x2 + y2 - 8x - 10y + 5 = 0 Particular Cases Y C(h, k) X r=k (i) Equation of a circle touching the X - axis. X' O Let a circle with centre at C(h, k) touch the X-axis at A. Y' A Then radius r = AC = k. Equation of the circle on given by (x - h)2 + (y - k)2 = k2 Example 3. Find the equation of a circle with centre (4, 5) and touches the X - axis. Solution: Here, centre (h, k) = (4, 5) Y Since the circle touches X- axis. Then r = k = 5 X' O C(4, 5) Now, equation of the circle is, Y' r=5 (x - h)2 + (y - k)2 = k2 X or, (x - 4)2 + (y - 5)2 = 52 or, x2 - 8x + 16 + y2 - 10y + 25 = 25 ? x2 + y2 - 8x - 10y + 16 = 0 Y which is the required equation of the circle. (ii) Equation of a circle touching the Y-axis. B hC Let a circle with centre c(h, k) touch the Y-axes at B. k Then radius r = BC = h. X' O A X Now, equation of the circle is (x - h)2 + (y - k)2 = h2 Y' Example 4. Find the equation of the circle with centre (-2, -4) and touching the Y-axis. Solution: Here, centre (h, k) = (-2 -4) Y Since the circle touches Y-axis r = h = -2 X' OX Now, equation of the circle is, (x - h)2 + (y - k)2 = h2 (-2, -4) or, (x + 2) + (y + 4)2 = (-2)2 or, x2 + 4x + 4 + y2 + 8y + 16 = 4 Y' ? x2 + y2 + 4x + 8y + 16 = 0 is the required equation. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187

vedanta Excel In Opt. Mathematics - Book 10 (iii) Equation of a circle touching both of the axes X and Y. Y Let a circle with centre (h, k) touch both of the axes X and Y at A and B. B h Then AC = BC = h = k = r C Then equation of the circle is X' k X (x - h)2 + (y - h)2 = h2 OA or, (x - k)2 + (y - k)2 = k2. Y' Example 5. Find the equation of the circle with centre (3,3) Y Solution: touching both of the axes. B 3 C(3, 3) Here, centre (h, k) = (3, 3) 3 Since the circle touch both of the axis, X' OA X h=k=r=3 Y' Now, equation of the circle is (x - h)2 + (y -h)2 =h2 or, (x - 3)2 + (y - 3)2 = 32 or, x2 - 6x + 9 + y2 - 6y + 9 =9 ? x2 + y2 - 6x - 6y + 9 = 0 is the required equation of the circle. (C) Equation of a circle in diameter form Let A(x1, y1) and B(x2, y2) be the P(x,y) ends of a diameter P(x, y) AB of a circle and O is the its centre. Let P(x,y) be any point on the circle. Join AP and BP A B Then by plane geometry, (x1, y1) (x2, y2) ‘APB = 90° [ Angle at semi-circle is right angle.] Now, slope of AP = (m1) = y - y1 slope of BP = m2 = y -xy-2x2 x- x2 Since AP is perpendicular to PB, we write, m1.m2 = - 1 or, y - y1 . y - y2 =-1 x- x1 x- x2 or, (y - y1) (y - y2) = - (x - y1) (x - x2) or, (x - x1) (x - x2) + (y - y1) (y - y2) = 0 which is the required equation circle in diameter form. The following properties of circles are to be noted (i) The straight line joining the centre of the circle and middle point of a chord is perpendicular to the chord. 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (ii) The perpendicular drawn from the centre of the circle on the chord bisects it. (iii) The angle at semi-circle is a right angle. (iv) A straight line is tangent to the circle if the length of the perpendicular drawn from the centre to the line is equal to the radius of the circle. (v) Two circles touch externally if the distance between the centres of two circles is equal to the sum of radii of the circles. Two circles touch internally if the distance between the centres of two circles is equal to difference of their radii. r1 r2 P C1 C1 C2 C2 Externally touched circles Internally touched circles r1 + r2 = C1 C2 PC1 = r1, PC2 = r2 r1 – r2 = c1 c2 Example 6. Find the equation of a circle whose ends of diameter are (-2, -3) and (2,3). Solution: The ends of a diameter are (-2. -3) and (2,3). i.e. (x1, y1) = (-2, -3), (x2, y2) = (2, 3) Now, equation of circle is given by, (x - x1) (x - x2) + (y - y1) (y - y2) = 0 or, (x + 2) (x - 2) + (y + 3) (y - 3) = 0 or, x2 - 4 + y2 - 9 = 0 ? x2 + y2 = 13 is the required equation of circle. (d) General Equation of the circle Consider the equation x2 + y2 + 2gx + 2fy + c = 0.......... (i) In which the coefficients of x2 and y2 are equal each having unity and there is no term containing xy. The above equation (i) can be written as x2 + 2gx + y2 + 2fy = -c or, x2 + 2gx + g2 + y2 + 2fy + f2 = g2 + f2 - c or, (x + g)2 + (y + f)2 = ( g2 + f2 - c )2 .......... (ii) which is in the form of (x - h)2 + (y - k)2 = r2 ........... (iii) Hence, the above equation of (i) is equation of circle. Comparing equation (ii) to (iii) Centre = (h, k) = (-g, -f) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189

vedanta Excel In Opt. Mathematics - Book 10 radius (r) = g2 + f2 - c The above equation (i) is called general equation of circle. Worked out Examples Example 1. Find the equation of circle with centre (4,1) and radius 5. Solution: Here, centre (h, k) = (4, 1) Example 2. radius (r) = 5 Solution: Now, equation of the circle is Example 3. (x -h)2 + (y - k)2 = r2 Solution: or, (x - 4)2 + (y - 1)2 = 52 or, x2 - 8x + 16 + y2 - 2y + 1 = 25 190 ? x2 + y2 - 8x - 2y = 8 is the required equation. Find the centre and radius of the circle x2 + y2 + 4x - 6y + 4 = 0 The given equation is x2 + y2 + 4x - 6y + 4 = 0 or, x2 + 4x + 4 + y2 - 6y + 9 = 4 + 9 - 4 or, (x + 2)2 + (y - 3)2 = 32 which is in the form of (x-h)2 + (y - k)2 = r2, ? centre (h, k) = (-2, 3), radius (r) = 3 units. If the equation of circle 4x2 + 4y2 - 24x - 20y - 28 = 0, find the coordinates of the centre and diameter of the circle. Equation of given circle is 4x2 + 4y2 - 24x - 20y - 28 = 0 or, x2 + y2 - 6x - 5y - 7 = 0 or, 5 52 (= 7 + 32 + 5 (2 2 2 2 ( (or, 25 x2 - 2.3.x + 32 + y2 - 2.y.( + + (x - 3)2 + 9+ 4 y- 5 ((2 2 =7 ( (or, y- 5 ( 2 89 (x - 3)2 + 2 (=4 ( (or, y- 5 2 89 2 (x - 3)2 + 2 = 2 which is in the form of (x - h)2 + (y - k)2 = r2 (centre (h, k) = 3, 5 and radius (r) = 89 units. 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 4. Find the equation of a circle whose ends of diameter are (2,4) and (4, 5) Solution: The ends of diameter are (x1y1) = (2, 4) and (x2,y2) = (4, 5) Now, equation of circle in diameter form is (x - x1) (x - x2) + (y - y1) (y - y2) = 0 or, (x - 2)(x - 4) + (y -4)(y - 5) = 0 or, x2 - 6x + 8 + y2 - 9y + 20 = 0 ? x2 + y2 - 6x - 9y + 28 = 0 is the required equation. Example 5. Find the equation of the circle whose centre is the point of intersection of Solution: x + 2y - 1 = 0 and 2x - y - 7 =0 and which passes through the point (3, 1).  Given equation are  x + 2y - 1 = 0 ...... (i) 2x - y - 7 = 0 ....... (ii) We get from equation (i), (3, 1) r= 2x–y–7=0 x = 1 - 2y .......... (iii) P C x+y–1=0 Putting the value of x in equation (ii), 2(1 - 2y) - y - 7 = 0 or, 2 - 4y - y - 7 = 0 y=-1 Putting the value of y in equation (iii), we get, x = 1 - 2(-1) = 3 ? centre of circle is C (3, – 1) Radius = CP = (3 – 3)2 + (1 + 1)2 = 2 units. Now, equation of circle is, (x – h)2 + (y – k)2 = r2 or, (x – 3)2 + (y + 1)2 = 22 or, x2 – 6x + 9 + y2 + 2y + 1 = 4 ? x2 + y2 – 6x + 2y + 6 = 0 is the required equation. Example 6. Find the equation of a circle with centre (6, 5) and touching the x-axis. Solution: Here, centre of the circle = (h, k) = (6, 5) Y Since the circle touches the X-axis C(6, 5) Radius = r = k = 5 r=5 Now, the equation of circle is X' O X (x – h)2 + (y – k)2 = k2 or, (x – 6)2 + (y – 5)2 = 52 Y' Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191

vedanta Excel In Opt. Mathematics - Book 10  or, x2 – 12x + 36 + y2 – 10y + 25 = 25 Example 7. or, x2 + y2 – 12x – 10x + 36 = 0 Solution: ? x2 + y2 – 12x – 10y + 36 = 0 is the required equation of the circle.  Find the equation of a circle passing through the origin and making Example 8. intercepts 4 and 6 on positive x and y axes. Solution: Let X-intercept OA=4 and Y-intercept OB=6 units of given circle. Let C be the centre of the circle, draw CM perpendicular to OA and CN perpendicular to OB. Y Then,ON = 1 OB = 1 × 6 = 3 B 2 2 1 1 OM = 2 ×OA = 2 ×4 = 2 N C (2, 3) Then centre of circle has coordinates (2,3) 3 X' 2 Now, radius of circle (r) = OC = 22 + 32 A X OM = 4 + 9 = 13 Y' Equation of required circle is (x - h)2 + (y - k)2 = r2 or, (x - 2)2 + (y - 3)2 = ( 13 )2 or, x2 - 4x + 4 + y2 - 6y + 9 = 13 or, x2 + y2 - 4x - 6y = 0 ? x2 + y2 - 4x - 6y = 0 is the required equation. Find the equation of a circle passing through the points (1,2), (3,4) and (5,2). Let, C(h,k) be the centre of circle which passes through the points M(1,2), N(5,2) and P (3,4) P (3, 4) Then CM = CN = CP CM2 = CN2 = CP2 (radii of same circle) M (1, 2) C (h, k) N (5, 2) CM2 = (1 - h)2 + (2 - k)2 = 1 - 2h + h2 + 4 - 4k + k2 = h2 + k2 - 2h - 4k + 5 ...... (i) CN2 = (5-h)2 + (2 - k)2 = 25 - 10h + h2 + 4 - 4k + k2 = h2 + k2 - 10h - 4k + 29 ..... (ii) CP2 = (3 - h)2 + (4 - k)2 = 9 - 6h + h2 + 16 - 8k + k2 = h2 + k2 - 6h - 8k + 25 ...... (iii) 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Taking CM2 = CN2 [from (i) and (ii)] h2 + k2 - 2h - 4k + 5 = h2 + k2 - 10h - 4k + 29 or, 8h = 24 ? h=3 Again, taking CN2 = CP2 or, h2 + k2 - 10h - 4k + 29 = h2 + k2 - 6h - 8k + 25 or, - 4h + 4k = - 4 or, h - k = 1 putting the value of h in this equation, k = 2 centre (h, k) = (3, 2) and radius (r) = Distance between C(3,2) and M(1,2) = (1-3)2 + (2-2)2 = 2 units. Now, equation of the required circle is (x-h)2 + (y-k)2 = r2 or, (x - 3)2 + (y -2)2 = 22 or, x2 - 6x + 9 + y2 - 4y + 4 = 4  ? x2 + y2 - 6x - 4y + 9 =0 which is required equation. Example 9. Find the equation of circle touching coordinate axes at (6,0) and (0,6) Solution: The circle touches the X-axis at A(6,0). Y-axis at B(0,6). Draw CA and CB perpendicular on X and Y-axis respectively. Then OACB will be a square. OA = BC = 6 Y OB = AC = 6 centre of the circle is C(6,6) and radius r = 6. B (0, 6) C (6, 6) Now, equation of the circle is given by (x - h)2 + (y - k)2 = r2 X' O X or, (x - 6)2 + (y - 6)2 = 62 A (6, 0) or, x2 - 12x + 36 + y2 - 12y + 36 = 36 Y' ? x2 + y2 - 12x - 12y + 36 = 0 is the required equation. Example 10. Find the equation of a circle which passes through the points (6,5) and (4,1) and with centre on the line 4x + y = 16 Solution: Let C(h, k)be the centre of the circle. The centre (h,k) lies on the line 4x + y = 16 4h + k = 16 ...... (i) Also, let equation of circle be (x-h)2 + (y-k)2 = r2 ....... (ii) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193

vedanta Excel In Opt. Mathematics - Book 10 The circle (ii) passes through the points P(6,5) and Q (4,1) Q (4, 1) P (6, 5) or, (h - 6)2 + (k - 5)2 = (h - 4)2 + (k + 1)2 or, -4h - 8k = -44 or, h + 2k = 11.........(iii) C (h, k) solving equations (i) and (iii), we get h = 3 and k = 4 centre = C (3,4) Radius (r) = CP = (6-3)2 + (5-4)2 = 32 + 12 = 10 units Now, equation of the circle is (x -h)2 + (y - k)2 = r2 or, (x-3)2 + (y - 4)2 = 10 or, x2 - 6x + 9 + y2 - 8y + 16 = 10 ? x2 + y2 - 6x - 8y + 15 = 0 is the required equation. Example 11. A line y - x = 2 cuts a circle x2 + y2 + 2x = 0 at two distinct points. Find the coordinates of the points. Also, find the length of the chord. Solution: Given equation of the line is y - x = 2 ......... (i) The line (i) cuts the circle at two distinct points A and B. We solve equations (i) and (ii) to find the coordinates of A and B. From equation (i), y = x + 2 Put the value of y in equation (ii), we get x2 + (x+2)2 + 2x = 0 x2 + y2 + 2x = 0 or, x2 + x2 + 4x + 4 + 2x = 0 or, 2x2 + 6x + 4 = 0 or, x2 + 3x + 2 = 0 or, x2 + 2x + x + 2 =0 y–x=2 B or, x(x + 2) + 1(x + 2) = 0 A or, (x + 2) (x + 1) = 0 x = -1, -2 when x = -1, y = -1 + 2 = 1 when x = -2, y = -2 + 2 = 0 The coordinates of A and B are (-1, 1) and (-2, 0) respectively. Now, the length of chord AB is given by AB = (-2+1)2+(0 -1)2 = 1 + 1= 2 units. 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 12. Find the equation of a circle which touches the X-axis at the point (6,0) and passes through the points (3,3). Solution: Let C(h,k) be the centre of the circle. Since it touches X-axis at D(6,0).  radius (r) = k and h = 6. Y Now, the equation of circle is (x - h)2 + (y - k)2 = k2 or, (x - h)2 + (y - k)2 = k2 It passes through the point (6,6), we get C = (h, k) (3 - 6)2 + (3 - k)2 = k2 or, 9 + 9 - 6k + k2 = k2 (3, 3) ? k=3 X' O D (6, 0) X Therefore, the required equation is Y' (x - 6)2 + (y - 3)2 = 32 or, x2 - 12x + 36 + y2 - 6y + 9 = 9 ? x2 + y2 - 12x - 6y + 36 = 0 is the required equation of circle. Example 13. Show that two circles x2 + y2 = 100 and x2 + y2 - 24x - 10y + 160 = 0 touch each other externally. Solution: The equation of given circles are x2 + y2 = 100.......... (i) x2 + y2 - 24x - 10y + 160 = 0 .......... (ii) From equation (i), radius (r1) = 10 units r1 = 13 r2 = 3 centre C1 = (0, 0) C1 = (0, 0) C2 = (12, 5) From equation (ii), radius = g2 + f2 - c (r2) = (-12)2 + (-5)2 - 160 = 144 + 25 - 160 = 9 = 3 units Centre (C2) = (12, 5) Distance between the centres = (12 - 0)2 + (5 - 0)2 = 144 + 25 = 169 = 13 units Sum of the radii of the circle = 10 + 3 = 13 Since the distance between the centres of circles is equal to sum of the radii of circles, two circles touch externally. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195

vedanta Excel In Opt. Mathematics - Book 10 Example 14. Find the equation of the circle which touches the positive Y-axis at a distance of 4 units from the origin and cuts of an intercept 6 from the X-axis. Solution : Let C (h, k) be the centre of the circle touching Y-axis at N and ON = 4. From C, draw perpendiculars CM and CN on X-axis and Y-axis respectively. Then, ON = CM = 4 =k k=4 Y Also, AM = MB = 1 AB = 3, 2 From right angled triangle CMA, AC2 = AM2 + CM2 N C (h, k) = 9 + 16 = 25 4 X' O A 4 BX AC = radius (r) = 5 units 3M (h, k) = (5, 4) Now, equation of the circle is Y' (x - h)2 + (y - k)2 = r2 or, (x - 5)2 + (y - 4)2 = 52 or, x2 - 10x + 25 + y2 - 8y + 16 = 25 or, x2 + y2 - 10x - 8y + 16 = 0 ? x2 + y2 - 10x - 8y + 16 = 0 is the required equation. Exercise 8.4 Very Short Questions 1. (a) Write down the equation of a circle with centre at O(o,o) and radius R. (b) Write down the equation of a circle with centre (h,k) and radius r. (c) Write radius and centre of general circle x2 + y2 + 2gx + 2fy + c = 0 (d) Write the equation of a circle whose ends of a diameter are (x1, y1) and (x2, y2) . 2. Write the equation of the circle under the following conditions. (a) centre at O(0,0) and radium 5 units. (b) centre at (-2, -3) and radius 6 units. (c) ends of a diameter are (1,2) and (3,4). (d) ends of a diameter are (4,5) and (-2, -3). (e) ends of a diameter are (-a,0) and (a,0). 3. Write the centre and radius of the following circles. (a) (x - 2)2 + (y - 4)2 = 25 (b) (x + 3)2 + (y + 5)2 = 81 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (c) (x - 2)2 + (y + 7)2 = 36 (d) (x - a)2 + (y - b)2 = c2 4. Find the centre and radius of the following circles : (a) x2 + y2 = 81 (b) x2 + y2 + 6x + 4y - 12 = 0 (c) x2 + y2 + 2px - 2qy + p2 + q2 = m2 + n2 (d) 4x2 + 4y2 + 16x - 24y - 52 = 0 (e) 9x2 + 9y2 - 36x + 6y = 107 (f) x2 + y2 - 2axcosT -2aysinT = 0 Short Questions 5. (a) Find the equation a circle whose centre is the (4,5) and touches x-axis. (b) Find the equation of a circle with circle with centre (-3, 5) and touching x-axis. (c) Centre (1,3) and touching the y-axis. 6. Find the equations of circles under the following condition. (a) Centre at (4, -1) and through (-2, -3) (b) radius 5 units and touching positive x-axis and y-axis. (c) touching the coordinate axes at (a,0) and (a,0). Long Questions 7. (a) Find the equation of a circle whose centre is the point of intersection of x + 2y - 1 = 0 and 2x - y - 7 = 0 and passing through the point (6,4) (b) Find the equation of a circle with equations of two diameters are x + y = 14 and 2x - y = 4 and passing through the point (4, 5). 8. Determine the points of intersections of a straight line and circle in each of the following cases: (a) x + y = 3, x2 + y2 - 2x - 3 = 0 (b) 2x - y + 1 = 0, x2 + y2 = 2 (c) x + y = 2, x2 + y2 = 4 Also find the lengths of the intercepts (chords) in each case. 9. Find the centre and radius of circles. (a) passing through the points P(-4, -2), Q(2,6) and R(2, -2) (b) passing through the points P(2, -1), Q(2, 3) and R(4, 1) 10. Find the equations of circle. (a) passing through the points (5,7),(6,6) and (2, -2). (b) passing through the points (-6,5), (-3,-4) and (2,1) 11. (a) Find the equation of circle passing through the origin and the point (4,2) and centre lies on the lines x + y = 1. (b) Find the equation of the circle passing through the points (3,2) and (5,4) and centre lies on the line 3x - 2y = 1. 12. (a) Find the equation of the circle which touches the x-axis at (4,0) and cuts of an intercept of 6 units from the y-axis positively. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197

vedanta Excel In Opt. Mathematics - Book 10 (b) Find the equation of the circle which touches the y-axis at (0, 4) and cuts of an intercept 6 units from the positive part of x-axis. 13. (a) Prove that the points A(2, -4), B(3, -1), C(3, -3) and D(0,0) are concyclic. (b) Prove that the points P(3,3), Q(7,1), R(4,0) and S(6,4) are concyclic. 14. (a) Show that the two circles x2 + y2 = 36 and x2 + y2 - 12x - 16y + 84 = 0 touch externally. (b) Show that the two circles x2 + y2 = 81 and x2 + y2 - 6x - 8y + 9 = 0 touch internally. Project Work 15. Draw a circle of radius 5 units taking centre at (2,3). On a graph, find equation of the circle. Take any four points on the circle and show that the points satisfy the equation of the circle. Discuss the conclusion. 2. (a) x2 + y2 = 25 (b) x2 + y2 + 4x + 6y = 23 (c) x2 + y2 - 4x - 6y + 11 = 0 (d) x2 + y2 - 2x - 2y - 23 = 0 (e) x2 + y2 = a2 3. (a) (2, 4), 5 (b) (-3, -5), 9 (c) (2, -7), 6 (d) (a, b), c 4. (a) (0, 0), 9 (b) (-3, -2), 5 (c) (-p, q), m2 + n2 (d) (-2, 3), 26 (e) 2, - 1 , 4 (f) (acosT, asinT), a 3 (b) x2 + y2 + 6x - 10y + 9 = 0 5. (a) x2 + y2 - 8x - 10y + 16 = 0 (c) x2 + y2 - 2x - 6y + 9 = 0 6.(a) x2 + y2 - 8x + 2y = 23 (b) x2 + y2 - 10x - 10y + 25 = 0 (c) x2 + y2 - 2ax - 2ay + a2 = 0 7. (a) x2 + y2 - 6x + 2y = 24 (b) x2 + y2 - 12x - 16y + 87 = 0 8. (a) (1, 2), (3, 0), 2 2 (b) (-1, -1), 1 , 7 , 65 (c) (0, 2), (2, 0), 2 2 5 5 2 9. (a) (-1, 2), 5 (b) (2, 1), 2 10. (a) x2 + y2 - 4x - 6y - 12 = 0 (b) x2 + y2 + 6x - 2y = 15 11. (a) x2 + y2 - 8x + 6y = 0 (b) x2 + y2 - 6x - 8y + 21 = 0 12. (a) x2 + y2 - 8x - 10y + 16 = 0 (b) x2 + y2 - 10x - 8y + 16 = 0 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 9 Trigonometry 9.0 Review of Trigonometricy Ratios of Compound Angle Let A and B be any two angles, the sum (A+ B) and the difference (A - B) are known as compound angles. The following are the list of formula of compound angles. (i) sin (A + B) = sinA.cosB + cosA. sinB (ii) cos (A + B) = cosA. cosB - sinA. sinB (iii) tan (A + B) = tanA + tanB 1 - tanA tanB cotA.cotB-1 (iv) cot (A + B) = cotB + cotA (v) sin (A -B) = sinA. cosB - cosA. sinB (vi) cos (A - B) = cosA.cosB + cosA. sinB (vii) tan (A -B) = tanA - tanB 1 + tanA.tanB cotA.cotB + 1 (viii)cot (A-B) = cotB - cotA (ix) sin (A + B). sin (A - B) = sin2A - sin2B = cos2B - cos2A (x) cos(A + B).cos (A - B) = cos2A - sin2B = cos2B - sin2A (xi) tan(A + B). tan(A - B) = tan2A – tan2B 1 - tan2A.tan2B cot2A.cot2B-1 (xii) cot(A + B). cot (A - B) = cot2B - cot2A 9.1 Trigonometric Ratios of Multiple Angles Let A be any angle. Then, 2A, 3A, 4A, ... etc. are called multiple angles of A. In this sub-units, we discuss the trigonometric ratios of angles 2A and 3A. I. Trigonometric Ratios of 2A We know that sin (A + B) = sinA.cosB - cosA.sinB. (a) sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA + tan2A + cot2A Here, sin2A = sin(A + A) = sinA.cosA + cosA.sinA ? sin2A = 2sinA.cosA............ (i) sin2A 2sinA.cosA = 2sinA.cosA 1 sin2A + cos2A Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199

vedanta Excel In Opt. Mathematics - Book 10 2sinA.cosA cos2A = sinA +cosA = 2tanA tan2A + 1 cos2A 2tanA ? sin2A = 1 + tan2A ...............(ii) 2tanA 2 1 1 2cot2A 2cotA 1 + tan2A cotA cotA cot2A + 1 + cot2A Also, sin2A = = = . = 1 1 1 + cot2A 2cotA ? sin2A = 1 + cot2A ......... (iii) Combining above (i), (ii) and (iii), we get, sin2A = 2sinA.cosA = 1 2tanA = 1 2cotA + tan2A + cot2A 1-tan2A cot2A -1 (b) cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1 + tan2A = 1 + cot2A Here, cos(A + B) = cosA.cosB - sinA.sinB Now, cos2A = cos(A + A) = cosA.cosA - sinA.sinA ? cos2A = cos2A - sin2A ............ (i) cos2A = cos2A - sin2A = 1 - sin2A - sin2A ? cos2A = 1 - 2sin2A ............... (ii) or, 2sin2A = 1 - cos2A. ? cos2A = 2cos2A - 1 .............(iii) 2cos2A = 1 + cos2A Again, cos2A = cos2A - sin2A cos2A-sin2A cos2A = cos2A - sin2A = sin2A + cos2A = 1- tan2A sin2A + cos2A 1 + tan2A cos2A 1-tan2A ? cos2A = 1+tan2A ........... (iv) 1 - 1 cot2A - 1 cot2A cot2A + 1 and cos2A = = .......... (v) 1 1 + cot2A Combining (i), (ii), (iii), (iv), and (v), we get cos2A = cos2A - sin2A = 1 - 2sin2A = 2cos2A - 1 = 1- tan2A = cot2A -1 1 + tan2A cot2A + 1 2tanA (c) tan2A = 1 - tan2A Here, tan(A + A) = tanA + tanA = 2tanA 1 - tanA.tanA 1 - tan2A 2 tanA ? tan2A = 1 – tan2A 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (d) cot2A = cot2A - 1 2cotA cotA.cotB - 1 Here, cot(A + B) = cotB + cotA Now, cot2A = cot(A + A) = cotA.cotA -1 = cot2A - 1 cotA + cotA 2cotA cot2A - 1 ? cot2A = 2cotA II. Trigonometric Ratios of 3A. (a) sin3A = 3sinA - 4sin3A Now, sin3A = sin(A + 2A) = sinA.cos2A + cosA.sin2A = sinA (1 - 2sin2A) + cosA. 2sinA.cosA = sinA - 2sin3A + 2sinA.cos2A = sinA - 2 sin3A + 2sinA(1 - sin2A) = sinA - 2sin3A + 2sinA - 2sin3A = 3sinA - 4sin3A ? sin3A = 3sinA - 4sin3A Also, 4sin3A = 3sinA - sin3A (b) cos3A = 4cos3A - 3cosA Here, cos3A = cos(A + 2A) = cosA.cos2A - sinA.sin2A = cosA. (2cos2A - 1) - sinA.2sinA.cosA = 2cos3A - cosA -2sin2A.cosA = 2cos3A - cosA - 2(1 - cos2A) cosA = 2cos3A - cosA - 2cosA + 2cos3A = 4cos3A - 3cosA. Also, 4cos3A = 3cosA + cos3A (c) tan3A = 3tanA - tan3A 1 - 3tan2A Here, tan3A = tan (A +2A) = tanA + tan2A = tanA + 2tanA 1-tanA.tan2A 1-tan2A 1- tanA. tanA-tan3A +2tanA 2tanA 1-tan2A = 1 - tan2A = 3tanA - tan3A 1 - tan2A -2tan2A 1 - 3tan2A 1-tan2A ? tan3A = 3tanA - tan3A 1- 3tan2A Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 201

vedanta Excel In Opt. Mathematics - Book 10 (d) cot3A = cot3A - 3cotA 3cot2A - 1 Here, cot3A = cot(A + 2A) = cotA.cot2A - 1 = cotA. cot2A -1 - 1 cot2A + cotA 2cotA cot2A -1 + cotA 2cotA = cot3A - cotA - 2cotA . 2cotA 2cotA cot2A - 1+2cot2A cot3A- 3cotA = 3cot2A - 1 ? cot3A = cot3A - 3cotA 3cot2A - 1 III. Geometrical Interpretation of Trigonometric Ratios of 2A. Let a revolving line OP make an angle ‘XOP = 2A with an initial line OX. Now, taking O as the centre and OP as radius of a circle, circle PMN is drawn. MN is the diameter of the circle. OP, MP, PN are joined. PR is drawn perpendicular to MN. MPN is semi-circle with a diameter MN, ‘MPN = 90° (angle at semi-circle is right angle). ‘MNP = 1 ‘NOP = 1 × 2A = A 2 2 (Circumference angle is half central angle standing on the same arc PN) ‘OPM = ‘OMP = A (?∆MOP is an isosceles base angles are equal). ‘MPR = 90° - A, ‘MPN = 90°, ‘RPN = 90° - ‘MPR = 90° - 90° + A = A ? ‘RPN = A Y In right angled ∆ PRM, P sinA = PR , cosA = RM , tanA = PR X' A 2A A X PM PM MR M RN In right angled ∆MPN, O sinA = PN , cosA = PM MN MN Now, in right angled ∆ ORP, Y' (i) sin2A = PR = PR = 2PR = 2PR . PM = 2sinA.cosA OP MN PM MN 1 MN 2 OR 2OR OR + OR (ii) cos2A = OP = 2OP = 2OP = (ON - RN) + (RM - OM) 2OP OM - RN + RM - OM = 2OP = RM - RN MN RM RN RM PM RN PN = MN - MN = PM . MN – PN . MN = cosA.cosA - sinA.sinA = cos2A - sin2A 202 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 (iii) tan2A = PR OR PR PR = 2PR = 2PR = 2. RM = 2. RM 2OR RM-RN 1- RN 1- RN . PR RM PR MR 2tanA 2tanA = 1 - tanA.tanA = 1 - tan2A Worked Out Examples Example 1. If sinA = 3 , find sin2A, cos2A and tan2A. Solution: 5 3 Here, sinA = 5 cosA = 1 - sin2A = 1 - 9 = 4 25 5 3 5 3 tanA = sinA = 4 = 4 cosA 5 Now, sin2A = 2sinA. cosA = 2. 53 . 4 = 24 5 25 cos2A = cos2A - sin2A = 16 - 9 = 16 - 9 = 7 25 25 25 25 24 25 tan2A = sin2A = 7 = 24 cos2A 7 25 Example 2. Prove that, cotT = ± 1+cos2T Solution: 1-cos2T We have 2cos2T = 1 + cos2T............ (i) Example 3. Solution and 2sin2T = 1 - cos2T........... (ii) Dividing (i) by (ii), we get, 2cos2T = 1 + cos2T 2sin2T 1-cos2T 1 + cos2T or, cot2T = 1-cos2T ? cotT = ± 1+cos2T = RHS Proved. 1-cos2T 4 If sinA = 5 , then, find the values of sin3A, cos3A and tan3A. Here, sinA = 4 5 16 Now, cosA = 1 - sin2A = 1 - 25 = 3 5 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 203

vedanta Excel In Opt. Mathematics - Book 10 4 5 tanA = sinA = 3 = 4 cosA 3 5 sin3A = 3sinA - 4sin3A = 3. 4 - 4. 64 5 125 = 12 - 256 = 300 - 256 = 44 5 125 125 125 cos3A= 4cos3A - 3cosA =4 27 - 3. 3 = 108 - 9 = 108-225 =- 117 125 5 125 5 125 125 3tanA - tan3A tan3A = 1 - 3tan2A 3. 4 - 64 4– 64 108 - 64 44 3 27 1- 27 7(3 - 16) 117 = = 16 = = – 16 3 1- 3. 9 Example 4. (If = 1 m + 1 (, then, show that : Solution: sinA 2 m (( ((a) 1 m2 (+1 ((b) = - 1 m3 + 1 cos2A = - 2 (m2 sin3A 2 m3 (( (Here, sinA = 1 m + ((1 2 m ( (a) cos2A = 1 - 2sin2A ( (= 1 - 2 . ( 1 m + 1 2 4 m (= ( 1 1 1 ((1 - 2. 4 m2 + 2.m. m + m2 =( 1- 1 m2 - 1 - 1 . 1 2 2 m2 (= - 1 m2 + 1 = LHS. Proved. 2 m2 (b) sin3A = 3sinA - 4sin3A (= 3 1 m+ 1 (- 4 .1 m + 1 3 2 m 8 m [ ( ]3 - (= 1 m+ 1 m + 1 2 2 m m [ ]3 (= 1 m+ 1 - m2 - 2.m. 1 - 1 2 m m m2 [ ]- (= 1 m+ 1 (m2 - 1 + 1 2 m m2 ( (= – 1 m + 1 m2 - m. 1 + 1 2 m m m2 (= – 1 m3 + 1 = RHS Proved. 2 m3 204 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 Example 5. Prove the following: Solution: (a) sin2A - sinA = tanA (b) 1- cosA = tan2A 1-cosA + cos2A 1+ cos2A (d) 1 + sec2A = cotA (c) tanA + cotA = 2cosec2A tan2A (a) LHS = sin2A - sinA 1-cosA + cos2A 2sinA.cosA - sinA sinA(2cosA - 1) = 1 - cosA + 2cos2A - 1 = cosA(2cosA -1) = sinA = tanA = RHS. proved cosA 1 - cos2A 2sin2A sin2A (b) LHS = 1 + cos2A = 2cos2A = cos2A = tan2A = RHS Proved. (c) LHS = tanA + cotA = sinA + cosA cosA sinA sin2A + cos2A 1 2 = sinA.cosA = 2. 2sinA.cosA = sin2A = 2cosec2A = RHS Proved. (d) LHS = 1 +sec2A = 1+  = cos2A + 1 × cos2A tan2A cos2A cos2A sin2A sin2A cos2A 2cos2A cosA = 2sinA.cosA = sinA = cotA = RHS Proved. Example 6. Proved that: Solution: (a) sin6T + cos6T= 1 (1 + 3cos22T) (b) cos4T = 1 - 8sin2T + 8sin4T 4 (a) LHS = sin6T + cos6T = (sin2T)3 + (cos2T)3 = (sin2T + cos2T) (sin4T - sin2T.cos2T + cos4T) = 1.(sin4T + cos4T - sin2T.cos2T) = (sin2T + cos2T)2 - 2sin2T.cos2T - sin2T.cos2T = 12 - 3sin2T.cos2T =1- 3 (2sinT . cosT)2 4 3 =1- 4 sin22T = 1 (4 - 3sin22T) 4 1 = 4 (1 + 3 - 3sin22T) = 1 {1 + 3 (1 - sin22T)} = 1 (1 + 3cos22T) = RHS. Proved. 4 4 (b) LHS = cos4T = cos2(2T) = 1 -2sin22T Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 205

vedanta Excel In Opt. Mathematics - Book 10 = 1 - 2[2sinT .cosT]2 = 1 – 8sin2T(1 - sin2T)] = 1 - 8sin2T + 8sin4T = RHS. Proved. Example 7. PLHroSved=thsaitn110s°in1-10ct°aons-6100c°°os130° =4 Solution: = 1 - sin60° Example 8. sin10° cos60°.cos10° Solution: cos60°.cos10° - sin60°.sin10° = sin10°.cos60°.cos10° Example 9. Solution: = cos (60° + 10°) = 2 . 2cos70° = 4cos70° 2 sin10°.cos10° sin20° 206 sin10° . 1 cos10° 2 4cos70° 4cos70° = sin(90° -70°) = cos70° = 4 = RHS. Proved. Proved that (a) cosec2T + cot4T = cotT - cosec4T (b) 4cosec2T . cot2T =cosec2T - sec2T (a) LHS = cosec2T + cot4T = 2ssiicnn11o22sTT2T+++cs2coisnosi4sn4c2TT2o(T2s.4TcT)os2T = 2sin2T . cos2T = = 2cos2T + 2cos22T - 1 2sin2T.cos2T 2cos2T(1 + cos2T) - 1 = 2sin2T.cos2T = 2cos2T.2cos2T - 1 = 2cos2T – 1 = 2sin2T.cos2T sin2(2T) 2sinT.cosT sin4T cosT 1 sinT - sin4T = cotT - cosec4T = RHS. Proved. (b) LHS = 4cosec2T.cot2T = 4. 1 . cos2T sin2T sin2T 4cos2T = 2sinT.cosT.2sinT.cosT = cos2T - sin2T = cos2T - sin2T sin2T.cos2T sin2T.cos2T sin2T.cos2T 1 = 1 - cos2T = cosec2T - sec2T = RHS. Proved. sin2T Proved that (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1) = 2cos8T + 1 LHS = (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1) = {(2cosT)2 - 12} (2cos2T - 1) (2cos4T -1) = (4cos2T –1) (2cos2T - 1) (2cos4T - 1) = {2(2cos2T - 1) + 1} (2cos2T - 1) (2cos4T - 1) Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = (2cos2T + 1) (2cos2T - 1) (2cos4T - 1) = (4cos22T - 1) (2cos4T - 1) = {2 (2cos22T - 1) + 1} (2cos4T - 1) = (2cos4T + 1) (2cos4T – 1) = 4cos24T - 1 = 2(2cos24T - 1) + 1 = 2cos8T + 1 = RHS. Proved. Example 10 Prove that cos2D + sin2D . cos2E = cos2E + sin2E . cos2D Solution: LHS = cos2D + sin2D . cos2E   = cos2D + sin2D(1 - 2sin2E) = cos2D + sin2D - 2sin2E . sin2D = 1 - 2sin2E . sin2D = cos2E + sin2E - 2sin2E . sin2D = cos2E + sin2E(1 - 2sin2D) = cos2E + sin2E . cos2D = RHS Proved. Example 11. Prove that 2+ 2+ 2+ 2 + 2cos16T = 2cosT Solution: LHS = 2+ 2+ 2+ 2 + 2cos16T = 2+ 2+ 2+ 2(1+cos16T) = 2+ 2+ 2+ 2.2cos28T = 2+ 2+ 2 + 2cos8T = 2+ 2+ 2(1+cos8T) = 2+ 2+ 2 . 2cos24T = 2+ 2+2cos4T = 2+ 2(1+cos4T) = 2+ 2.2cos22T) = 2(1+cos2T) = 2.2cos2T = 2cosT = RHS. Proved. Example 12. Prove that 1 - 1 cotA = cot2A tan3A - tanA cot3A - 1 1 Solution: LHS = tan3A - tanA - cot3A - cotA = 1 - 1 tan3A - tanA 1 - 1 tan3A tanA = 1 tanA - tanA.tan3A tan3A - tanA - tan3A Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 207

vedanta Excel In Opt. Mathematics - Book 10 = 1 + tanA.tan3A = 1= 1 tan3A - tanA tan3A - tanA tan(3A - A) = 1 = cot2A 1 +tanA.tan3A tan2A = RHS. Proved. Example 13. Proved that 4(cos320° + cos340°) = 3(cos20° + cos40°) Solution: LHS = 4 (cos320 + cos340°) = 4cos320 + 4cos340° = 3cos20° + cos(3.20°) + 3cos40° + cos(3.40°) = 3cos20° + cos60° +3cos40° + cos120° = 3(cos20° + cos40°) + 1 - 1 2 2 = 3(cos20° + cos40°) = RHS. Proved. Example 14. Prove that sin3T + sin3T = cotT Solution: cos3T - cos3T sin3T + sin3T 3sinT - 4sin3T + sin3T LHS = cos3T - cos3T = cos3T - 4cos2T + 3cosT = 3sinT - 3sin3T = 3sinT(1-sin2T) = sinT . cos2T 3cosT - 3cos3T 3cosT(1-cos2T) cosT . sin2T cosT = sinT = cotT = RHS. Proved Example 15. Prove that : tanT + 2tan2T + 4tan4T + 8cot8T = cotT Solution: LHS = tanT + 2tan2T + 4tan4T + 8cot8T = tanT + 2tan2T + 4tan4T + 8 tan(2.4T) 8 = tanT + 2tan2T + 4tan4T + 2tan4T 1 - tan24T (= tanT + 2tan2T + 8(1-tan24T) 2tan4T 4tan4T + ( (= tanT + 2tan2T + 4 tan4T + 1-tan24T( tan4T( ( (= tanT + 2tan2T + 4 tan24T + 1-tan24T tan4T 4 = tanT + 2tan2T + tan4T = tanT + 2tanT 4 2tan2T (= tanT + 2 1 - tan22T 1-tan22T tan2T + tan2T = tanT + 2 tan22T +1- tan22T tan2T 208 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = tanT + 2 2tanT 1 - tan2T = tanT + 2(1 - tan2T) 2tanT tan2T + 1 - tan2T 1 = tanT = tanT = cotT = RHS. Proved. Example 16. Prove that (a) 4sin3A.cos3A + 4cos3A.sin3A = 3sin4A (b) cos3A.cos3A + sin3A. sin3A = cos32A Solution: (a) We have, 4sin3A = 3sinA - sin3A 4cos3A = 3cosA + cos3A Now, LHS = 4sin3A.cos3A + 4cos3A.sin3A = (3sinA – sin3A). cos3A + (3cosA + cos3A).sin3A =3sinA.cos3A - sin3A.cos3A + 3sin3A.cos3A + cos3A.sin3A = 3(sinA.cos3A + sin3A.cosA) = 3.sin(A + 3A) = 3sin4A = RHS Proved. (b) LHS = cos3A cos3A + sin3A.sin3A = 1 (3cosA + cos3A).cos3A + 1 (3sinA - sin3A).sin3A) 4 4 1 = 4 [3cosA.cos3A + cos23A + 3sinA.sin3A - sin23A] = 1 [3(cosA.cos3A + sinA.sin3A) + (cos23A - sin23A)] 4 1 = 4 [3cos(A - 3A) + cos6A] = 1 (3cos2A + cos6A) 4 1 1 = 4 [3cos2A + cos(3.2A)] = 4 [3cos2A + 4cos32A – 3cos2A] = 4cos32A 4 = cos32A =RHS. Proved. Example 17. Prove that cot(45° + T) + tan(45° - T) = 2cos2T 1 + sin2T Solution: LHS = cot(45° + T) + tan(45 - T) = cot45°.cotT - 1 + tan45° - tanT cot45° + cotT 1 +tan45°.tanT cotT - 1 1-tanT = 1 + cotT + 1 + tanT 1 -1 1 - tanT tanT 1 + tanT = + 1 + 1 tanT Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 209

vedanta Excel In Opt. Mathematics - Book 10 = 1-tanT + 1-tanT 1 + tanT 1 + tanT sinT 2(1-tanT) 1- cosT 1+tanT = = 2. 1 + sinT cosT (cosT - sinT) cosT - sinT cosT + sinT =2 cosT + sinT = 2. (cosT + sinT) × cosT + sinT = 2. cos2T = 2cos2T = RHS. Proved. sin2T + cos2T + 2sinT.cosT 1 + sin2T Example 18. Prove that 2sinT + 2sin3T + 2sin9T = tan27T - tanT cos3T cos9T cos27T 2sinT 2sin3T 2sin9T Solution: LHS = cos3T + cos9T + cos27T = 2sinT.cosT + 2sin3T.cos3T + 2sin9T.cos9T cos3T.cosT cos9T.cos3T cos27T.cos9T = sin2T + sin6T + sin18T cos3T.cosT cos9Tcos3T cos27T.cos9T = sin(3T - T) + sin(9T - 3T) + sin(27T - 9T) cos3T.cosT cos9T.cos3T cos27T.cos9T sin3T.ccoossT3T-.ccoossT3T.sinT+ sin9T.cos3T - cos9T.sin3T = cos9T.cos3T +sin27T.ccooss99TT.-cocos2s277TT.sin9T = tan3T - tanT + tan9T - tan3T + tan27T - tan9T = tan27T - tanT = RHS. Proved. Example 19. Without using table or calculator. Show that sin18° = 5-1 . Solution: Let T = 18° 4 Then 5T = 90° or, 2T + 3T = 90° or, 3T = 90° - 2T Taking cosine on both sides, we get cos3T = cos(90° - 2T) or, cos3T = sin2T or, 4cos3T - 3cosT = 2sinTcosT since sinT= sin18°≠ 0, dividing both sides by cosT. 4cos2T - 3 = 2sinT or, 4 - 4sin2T - 2sinT - 3 = 0 or, -4sin2T - 2sinT + 1 = 0 or, 4sin2T + 2sinT - 1 = 0 210 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 This equation is quadratic is sinT in the form of ax2 + bx + c = 0, where a = 4, b = 2, c = -1 x= b± b2 - 4ac 2a or, sinT = - 2 ± 22 - 4.4(-1) = 2± 4 + 16 2.4 8 = 2± 20 = 2±2 5 = -1 ± 5 8 8 4 since value of sin18° is positive. We take only positive sign. Hence, sinT = sin18° = 5 -1 . Proved. 4 ( ( (Prove that cosSc 2Sc 3Sc 1 Example 20. 7 (, cos 7 , cos 7 = 8 Solution: ( ( (LHS = cos Sc ( (( 2Sc 3Sc 7 ( (( , cos 7 , cos 7 2sin Sc . cos Sc . cos 2Sc . cos 3Sc = 7 7 7 7 Sc 2sin 7 2.sin 2Sc . cos 2Sc . cos 2Sc 7 7 7 = 2.2sin Sc 7 4Sc 3Sc sin 7 . cos 7 = 4sin Sc 7 3Sc 3Sc 2sin 7 . cos 7 [ ( ) ] sin 4Sc Sc = sin 3Sc = 4. 2sin Sc 7 = sin S - 7 7 7 (= sin 6Sc sin Sc - Sc sin Sc 1 7 7 7 8 = = = = RHS. Proved. 8sin Sc 8sin Sc 8sin Sc 7 7 7 Exercise 9.1 211 Very Short Questions 1. (a) Define multiple angles with an example. (b) Express sin2T in terms of sinT, cosT and tanT. (c) Express cos2T in terms of sinT, cosT and tanT. (d) Express cos2T in terms of cotT. Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 2. (a) Express sin3T in terms of sinT (b) Express cos3T in terms of cosT (c) Express tan3T in terms of tanT. (d) Express tanT in terms of sin2T and cos2T Short Questions 3. (a) If sinT = 1 , find the value of cos2T. 2 1 (b) If tanT  3 , find the value of tan3T (c) If tanT = 1 , find the value of tan3T. 2 4 4. (a) If sinT = 5 , find the value of sin2T, cos2T, tan2T. (b) If sinT = 1 , find the values of sin2T, cos2T, tan2T. 2 3 (c) If tanT = 4 , find the values of sin2T, cos2T and tan2T (d) If sinT = 3 , find the value of sin3T and cos3T. 2 1 7 5. (a) If sinT = 4 , show that cos2T = 8 (b) If cosT = 6 , show that cos2T = 11 52 25 (c) If tanT = 5 , show that tan2T = 120 12 119 6. By using the formula of cos2T, establish the following : (a) sinT = ± 1-cos2T (b) cosT = ± 1+cos2T 2 2 (c) tanT = ± 1-cos2$ 1+cos2A (7. 1 1 (a) If sinT = 2 p + P ((, show that : ( (( ( ( ((i) 1 1 (ii) 1 1 cos2T = – 2 p2 + P2 sin3T = – 2 p3 + P3 ((b) 1 1 , show that : If cosT = 2 p + P ((i) 1 1 ((ii) 1 1 cos2T = 2 p2 + P2 cos3T = 2 p3 + P3 8. Prove that : (a) 1 - cos2T = tanT (b) sin2T = cotT sin2T 1-cos2T sin2T 1 - cos2T (c) tanT = 1 + cos2T (d) 1 + cos2T = tan2T (e) sin2T = 2cotT 1 (f) 1 - sin2T = 1 - tanT cot2T + cos2T 1 +tanT cos2T 1 - tanT cosT cosT (g) 1 + sin2T = 1 + tanT (h) cosT - sinT - cosT + sinT = tan2T (i) 1 + sin2A = sinA + cosA (j) cotT- tanT = cos2T cos2A cosA - sinA cotT + tanT 4tanT cosT - sinT (k) tan2T + sin2T = 1 - tan4T (l) cosT + sinT = sec2T - tan2T 212 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 sin5T cos5T sin3T + cos3T 1 sinT cosT cosT + sinT 2 (m) - = 4 cos2T ((n) = 1 - sin2T 1 - sin2T Sc cos2T 4 (o) cos4T - sin4T = cos2T ( (p) = tan - T ( ((q) Sc cos2T = tan 4 - T 1 +sin2T 9. Prove that (a) sin2T - cosT = cotT (b) 1 + sin2T - cos2T = tanT 1 - sinT - cos2T 1 + sin2T + cos2T (c) sinT + sin2T = tanT (d) 1-cos2T + sin2T = tanT 1 + cosT + cos2T 1 + sin2T + cos2T (e) (sinT + cosT)2 - (sinT - cosT)2 = 2 sin2T (f) sinT + cosT + cosT- sinT = 2 sec2T cosT - sinT cosT + sinT (g) (1 + sin2T + cos2T)2 = 4cos2T(1 + sin2T) (h) 1 - 1 cotT = cotT tan2T - tanT cot2T - 10. Prove that following : ((a) cotT-1 2 (((b) 1 + tan2 Sc - T( cotT+1 4 ( 1-sin2T = ( = cosec2T 1+sin2T ( Sc - ((c) sin2 1 - tan2 4 T Sc 4 - T = 1 (1 - sin2T) 2 b 11. If tanT = a , prove that a cos2T + b sin2T = a. Long Questions 12. Prove that (a) cos4T + sin4T = 1 (3 + cos4T) (b) cos6T - sin6T = cos2T (1 - 1 sin22T) 4 4 1 1 (c) sin4T = 8 (3 - 4 cos2T + cos4T) (d) cos8T + sin8T = 1 - sin22T + 8 sin42T. (e) cos5T =16cos5T – 20cos3T + 5cosT (f) sin5T = 16sin5T - 20sin3T + 5sinT 13. Prove that (a) 3 + 1 =4 (b) cosec10° - 3 sec10° = 4 sin40° cos40° 14. Prove that (a) (2cosT + 1) (2cosT - 1) = 2cos2T + 1) (b) sec8T - 1 = tan8T sec4T - 1 tan2T (c) tanT + 2tan2T + 4tan4T + 8cot8T = cotT (d) sin2D - cos2D.cos2E = sin2E - cos2E.cos2D (e) 2+ 2+ 2 + 2cos8T = 2cosT Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 213

vedanta Excel In Opt. Mathematics - Book 10 (f) sin2D - sin2E = tan (D + E) sinD.cosD - sinE.cosE 15. Prove that (a) 4(cos310° + sin320°) = 3(cos10° + sin20°) (b) sin310° + cos320° = 3 (cos20° + sin10°) 4 (c) 4(cos320° + sin350°) = 3(cos20° + sin50°) ( ((d) Sc Sc tanA + tan ( ( ((3 + $ – tan 3 - A = 3tan3A (( ( ( 16. Prove that ((( ( (a) cot (A + 45°) - tan (A - 45°) =(( 2cos2A 1 + sin2A (b) tan(A + 45°) - tan (A - 45°) = 2sec2A. (c) tan(A + 45°) + tan (A - 45°) = 2tan2A. 17. (a) If 2tanD = 3tanE, then prove that, (i) tan (D - E) = sin2E (ii) tan(D + E) = 5sin2E 1 5 - cos2E 5cos2E - 1 1 (b) If tanT = 7 and tanE = 3 , prove that : cos2T = sin4E 18. Prove that (a) cosA - 1 + sin2A = tanA (b) 1 - 1 = cot4T sinA - 1 + sin2A tan3T + tanT cot3T + cotT  (c) cotA - tanA = 1. cotA - cot3A tan3A - tanA 19. Prove that ( ( ( ((a) 8 1 + sin Sc 1+sin 3Sc 1- sin 5Sc 1-sin 7Sc =1 8 8 8 8 ( ( ( ((b) 1-cos Sc 1- cos 3Sc 1- cos 5Sc 1- cos 7Sc = 1 8 8 8 8 8 ( ( ( ((c) sin4 Sc + sin4 3Sc + sin4 5Sc + sin4 7Sc = 3 8 8 8 8 2 20. Prove that cos233° – cos257° = 2 sin210.5° – sin234.5° 3 21. (a) Prove that cos3A + cos3(120° + A) + cos3(240° + A) = 4 cos3A (b) Prove that sin3A + sin3(60° + A) + sin3(240° + A) = – 3 sin3A 4 Project Work 22. Prepare a report by calculating the values of sin18°, cos18°, sin36°, cos36°, cos54°, tan18°, and tan54° by using relations of multiple angle ratios in trigonometry. For this project work, the students of class can be divided in groups as required. 3. (a) 1 (b) f (c) 11 4.(a) 24 , - 7 , - 24 2 2 25 25 7 3 1 24 7 24 (b) 2 , 2 , 3 (c) 25 , 25 , 7 (d) 0, -1 214 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 9.2 Trigonometric Ratios of Sub-multiple Angles Let A be any angles. Then A , A , A , ....... etc. are called sub-multiple angles of A. In this sub-unit, we discuss 2 3 4 ratios of sub-multiple angles $2 , $ the trigonometric 3 , etc. (I) Trigonometric Ratios of Half Angles : sinA = 2sin A A 2tan A2 = 2cot A ((a) 2 cos 2 = 2 = A A 1 + tan2 2 1 + cot2 2 A Here, sinA A + 2 ( 2 ( A A A A A A = sin 2 cos 2 + cos 2 . sin 2 = 2sin 2 cos 2 . ⸫ sinA = 2sin A cos A ............ (i) 2 2 A A sinA = 2sin 2 cos 2 2sin A . cos A 2 2 = A A sin2 2 + cos2 2 Dividing numerator and denominator by cos2 A 2 A A 2tan 2 2tan 2 = 1sh+owtanth2 aA2t ⸫ sinA = 1+ tan2 A ........... (ii) can 2 Similarly we 2cot A 2 sinA = A .......... (iii) 1+ cot2 2 Combing (i), (ii) and (iii), we get, A A 2tan A 2cot A 2 2 2 2 ⸫ sinA = 2sin . cos = A = A 1+ tan2 2 1+ cot2 2 A A 1-tan2 A cot2 A - 1 2 2 2 cot2 2 cosA - sin2 ((b) = cos2 = 1+ tan2 A = A + 2 21 Here, cosA = cos A + A 2 2 = cos A . cos A - sin A . sin A 2 2 2 2 A A = cos2 2 - sin2 2 cosA = cos2 A - sin2 A ........... (i) 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 215

vedanta Excel In Opt. Mathematics - Book 10 = cos2 A - 1 + cos2 A 2 2 A = 2cos2 2 - 1 Also, 2cos2 A = 1 + cosA 2 A A Again, cosA = cos2 2 - sin2 2 = 1 - 2sin2 A 2 A 2.sin2 2 = 1 - cosA cosA = cos2 A - sin2 A 2 2 A A cos2 2 - sin2 2 = sin2 A + cos2 A 2 2 Dividing numerator and denominator by cos2 A . 2 A 1-tan2 2 ⸫ cosA = 1+ tan2 A ....... (ii) 2 cot2 A - 1 Similarly, we can show cotA = 2 ............ (iii) A cot2 2 + 1 Combining (i), (ii) and (iii), we get A A 1-tan2 A cot2 A -1 2 2 2 2 ⸫ cosA = cos2 - sin2 = A = . 1+ tan2 2 A cot2 2 + 1 2tan2 A 2 tanA = ((c) 1- tan2 A 2 A A tan 2 + tan 2 A( + A Here, tanA = tan 2 ( 2 = 1- tan A . tan A 2 2 2tan A 2 ⸫ tanA = A 1- tan2 2 ((d) A + A cotA = cot 2 2 cot A . cot A - 1 cot2 A - 1 2 2 2 Here, cotA = A A = A cot 2 + cot 2 2cot 2 cot2 A - 1 2 ⸫ cotA = A 2cot 2 216 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 II. Trigonometric Ratios of A in terms of A . 3 ((a) A 2A ( A A sinA = sin 3 + 3 = 3sin 3 - 4sin3 3 ( (Here, sinA ( 2A A = sin 3 + 3 = sin 2A . cos A + cos 2A . sin A 3 3 3 3 (= A A A A ( A 2sin 3 . cos 3 cos 3 + 1- 2sin2 3 . sin 3 (= A 1- sin2 A A A 2sin 3 3 + sin 3 - 2sin3 3 = 2sin A - 2sin3 A + sin A - 2sin3 A 3 3 3 3 A A ? sinA = 3sin 3 - 4 sin3 3 = 4cos3 A - 3cos A ((b) cosA 3 3 Here, cosA = cos (2A+ A ( 3 3 = cos 2A (.cos A - sin 2A . sin A 3 ( 3 3 3 (= A A A A A 2cos2 3 -1 . cos 3 - 2sin 3 . cos 3 . sin 3 (= A - cos A - A 1 - cos2 A ( 2cos3 3 3 2cos 3 3 = 2cos3 A - cos A - 2cos A + 2cos3 A = 4cos3 A - 3cos A 3 3 3 3 3 3 A A ? cosA = 4cos3 3 - 3cos 3 ((c) tanA = tan A 3tan A - tan2 A 3 3 3 3 . = A 1 - 3tan2 3 (Here, tanA 2A A = tan 3 + 3 2tan A A 3 3 + 2tan tan 2A + tan A 1 - tan2 A 3 3 3 = = 1 - tan 2A . 3tan A A 3 3 1 - 2tan 3 . A 3 1 - tan2 A tan 3 A A A A A 2tan 3 + tan 3 - tan3 3 3tan 3 - tan3 3 = 1 - tan2 A - 2tan2 A = 1 - 3. tan2 A 3 3 3 A A 3tan 3 - tan3 3 ? tanA = 1 - 3. tan2 A 3 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217

vedanta Excel In Opt. Mathematics - Book 10 ((d) cotA = cot A ( cot3 A - 3cot A 3 3 3 3 . = A (For practice do yourself.) 3cot2 3 - 1 Worked out Examples Example 1. If sin T = 3 find the values of. Solution: 2 5 (a) sinT(b) cosT (c) tanT Example 2. Solution: Here, sin T = 3 2 5 cos T = 1 - sin2 T = 1 - 9 2 2 25 = 25-9 = 16 = 4 25 25 5 T 3 T sin 2 5 3 2 4 4 tan = cos T = 5 = 2 3 4 24 (a) sinT = 2 sin T . cos T = 2. 5 . 5 = 25 2 2 T T 16 9 16-9 7 (b) cosT = cos2 2 - sin2 2 = 25 - 25 = 25 = 25 2tan T 2. 3 3 16 3 16 24 2 4 2 16-9 2 7 7 (c) tanT = = 9 = × = × = 1 - tan2 T 1- 16 2 Alternative 24 tanT  sinT = 25 = 24 cosT 7 7 25 If cos45° = 1 , show that cos2212 °= 1 . 2+ 2 2 2 Here, cos45° = 1 2 (or, 45° =1 2 2 cos2 ( or, 224cc25oo°ss22is442255p°°osi-=t1iv=e12[( 1 ( cosT = 2cos2 T - 1) or, 2 quadrant] 2 of cos + 45° 1 2 lies ( (Value in 1st ? cos 45° = 1 + 1 = 2+1 × 2 ( 2 22 2 22 2 218 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 = 2+ 2 = 1 . 2+ 2 2 2 ? cos 45° = 1 . 2 + 2 Proved. 2 2 Example 3. Prove that Solution: (a) tan T = sinT (b) 1 + sinT - cosT = tan T Example 4. 2 1 + cosT 1 + sinT + cosT 2 Solution: sinT (a) RHS = 1 + cosT 2sin T . cos T sin T T 2 2 2 2 = = = tan = LHS. Proved. 2cos2 T cos T 2 2 1 + sinT - cosT 1 + sinT + cosT ((b) LHS = 1 + (2sin T . cos T - 1- 2sin2 T (( 2 2 2 = ( T T T 1 + 2sin 2 . cos 2 + 2cos2 2 -1 ( 1 + (2sinT . cos T - 1+ 2sin2 T (= ( 2 2 2 ((2cosT sin T + cos T 2 2 2 2sin T cos T + sin T (= 2 2 2 = RHS. Proved. T T T 2cos 2 cos 2 + sin 2 Prove that ((a) tan Sc + A = secA + tanA 4 2 ( ((b) cos2 Sc D Sc D D 4 - 4 - sin2 4 - 4 = sin 2 ((a) Sc + A LHS = 4 2 tan Sc + tan A 1+ sin A 4 2 2 = = Sc $ $ 1 - tan 4 . tan 2 1 - tan 2 1 + sin A 2 cos A cos A + sin A cos $ 2 2 2 2 = = × $ $ A A 1 - sin 2 cos 2 cos 2 + sin 2 cos A 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 219

vedanta Excel In Opt. Mathematics - Book 10 cos $ + sin $ cos $ + sin $ 2 2 2 2 = × A A A A cos 2 - sin 2 cos 2 + sin 2 cos2 $ + sin2 $ + 2sin $ .cos $ 1 + sinA 2 2 2 2 cosA = = A A cos2 2 - sin2 2 = 1 + sinA = secA + tanA = RHS. Proved. cosA cosA ( ((b) LHS = cos2 Sc - D (- sin2Sc-D ( (= cos2 4 4 (( 4 4 (( Sc - D = cos Sc - D = sin D = RHS Proved. (Prove that (cosD + cosE)2 + (sinD + sinE)2 = 4cos244 2 2 2 Example 5: D-E Solution: 2 LHS = (cosD + cosE)2 + (sinD + sinE)2 = cos2D + 2cosD cosE + cos2E + sin2D + 2sinD.cosD + sin2E = (cos2D + sin2D) + (sin2E + cos2E) + 2(cosD.cosE + sinD.sin2E) (= 2[1 + cos(D - E)] = 2.2cos2 = 1 + 1 + 2cos (D - E) (= 4cos2 D-E (( (( D-E 2 (( ( 2 = RHS. Proved. ( (( Example 6: (Proved that (( ( ( (1+ cos3Sc 1 + cos Sc ((   1+ cos5Sc 1+ cos7Sc = 1 ((   8 Solution: Here, ( ( { ( )}{ ( )}LHS. = ( ( ( ( (= 1+ cos Sc 1+ cos3Sc 1 + cos S - 3Sc 1 + cos Sc - Sc ( (=   8 8 ( { ( )}= ( (= 1+ cos Sc 1+ cos3Sc 1- cos3Sc 1- cosSc     1- cos2 Sc 1- cos23Sc   1- cos2 Sc 1 - cos2 Sc - Sc  2 8 1- cos2Sc 1- sin2Sc  = sin2 Sc. cos2 Sc  ( (= 1  2 2sinSc. cos Sc  ( (=1  2 sin2Sc  220 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 ( ( (= 1 sin2Sc = 1 1 = 1 = RHS Proved. (    ( (Example 7. ( ( ( (Solution: D 1 1 1 a2 +a12 If cos 2 = 2 a + a , prove that cosD = 2 Here, cos D = 1 a + 1 ,   a cosD = 2cos2 D - 1 [ ( (]= 2  1 a + 1 2 -1 ( (=  a 2. 1 a2 + 2.a 1 +a12 -1 ( (= 1  a ( ( a2 + 1 +1-1= 1 a2 + 1 = RHS. Proved. a2  a2 Exercise 9.2 Very Short Questions 1. (a) Define sub-multiple angle with an example. (b) Express sinT in terms of sin T and cos T .   T T T (c) Express cosT in terms of cos  , sin  and tan  . (d) Express tanT in terms of tan T .  T 2. (a) Express sinT in terms of sin  . (b) Express cosTin terms of cos T .  T (c) Express tanT in terms of tan  . 3. (a) If sin T =  , find the value of sinT.   T 3 (b) If cos  = 2 , find the value of cosT. (c) If tan T = 3, find the value of tanT. (d) If  = find the value of cosT. T 4 cos  5 , Short Questions 4. (a) If cos30° = 23, find the values sin15°, cos15° and tan15°. 1 (b) If cos45° = 2 , then prove that : (i) sin2212° = 1 2- 2 (ii) cos2212° = 1 2+ 2 (iii) tan2212° = 3-2 2 2 2 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 221

vedanta Excel In Opt. Mathematics - Book 10 (c) If cos330° = 3 prove the following : 2 3-1 3-1 (i) cos165° = - 22 (ii) sin165° = 22 (iii) tan165° = - (2 - 3) 5. Prove that: ( ( ( ((a) 1 - tan2 x (b) sinx = 3 sin x - 4sin3 x 2  cosx = x 1 + tan2 2 ( ( ( ( ( ( ( (3tan x - tan3 x cot3 x - 3cot x   1 - 3tan2 x 3cot2 x - 1 ( ( ( ((c) tanx = (d) cotx = ( ( ( (6.   (a) If cos T = 1 p + 1 , prove that cosT = 1 p2 + 1 . ( ( ( ((b)  2 p 2 p2 If cos T = 1 p + 1 , prove that cosT = 1 p3 + 1 . ( ( ( ((c)  2 p 2 p3 If sin T = 1 p+ 1 , prove that cosT = - 1 p2 + 1 .  2 p 2 p2 (7. Prove that: ((b) 1 + sinT = (a) 1 - sinT = sin T - cos T ((2 sin T +cos T 2 2 2 2 2 (c) 1 + cosT = cot T (d) 1 + secT = cot T sinT 2 tanT 2 1 + cosT (e) cosecT - cotT =tan T (f) 1 - cosT = cot2 (2T) 2 A 1 - sinA 1 - tan 2 sin3 T - cos3 T  1 cosA 2 2 2 (g) = A (h) = 1 + sinT 1 + tan 2 sin T - cos T 2 2 1+ secT T 1+ sinT cos T + sin T  secT 2 cosT 2 2 (i) = 2cos2 (j) = cos T - sin T (((l) 2 2 ((k) (( Sc T 1 - tan2 Sc - T ( T (4 2 4 4 2 1 - 2sin2 - = sinT ( = sin Sc T ( 1 +tan2 4 – 4 Long Questions ( (8. Prove that:T T 2sinT - sin2T = tan2 T (a) cos4 2 - sin4 2 = cosT 2sinT + sin2T 2 ((b) (c) 1 sin2T . cosT = tan T (d) 1 +sinT = tan2 Sc + T + cos2T 1 + cosT 2 1 - sinT 4 2 cos T  1+sinT T sin T  1+sinT T 2 2 2 2 (e) = tan (f) = cot T T sin 2  1+sinT cos 2  1+sinT 222 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

vedanta Excel In Opt. Mathematics - Book 10 9. Prove that ((a) tan Sc + T(((( = secT + tanT ((b) 4 2( ((( Sc 1 - sinT tan 4 – T (=1 + sinT ( ((c) 2 ( sec Sc + T(( . sec Sc – T = 2secT ((d) 4 2(( 4 2 tan Sc – T( = cosT ( ((e) 4 2(( 1 +sinT 2cosT cot T + Sc ( – tan T – Sc = 1 +sinT ( ((f) 2 4 ( 2 4 tan Sc + T + tan Sc - T = 2secT 4 2 4 2 10. Prove that ((a) (cosD - cosE)2 + (sinD - sinE)2 = 4sin2 ((b) (sinD + sinE)2 + (cosD + cosE)2 = 4cos2 D-E 2 D-E 2 11. Prove that ( ( ( ((a) cos 2Sc . cos 4Sc . cos 8Sc . cos 16Sc = 1 ( ( ( ((b) 15 15 15 15 16 1+cos Sc 1+cos 3Sc 1+cos 5Sc 1+cos 7Sc = 1 8 8 8 9 8 Project Work 12. Discuss how to find the value of tan 7 1° . Then show that 2 1° tan 7 2 = 6– 3+ 2 – 2. 3. (a) 3 (b) 1 (c) - 3 (d) - 44 2 2 125 3 -1 3 + 1 4. (a) 2 2 , 2 2 , 2 – 3 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 223

vedanta Excel In Opt. Mathematics - Book 10 9.3 Transformation of Trigonometric Formulae The sum or difference forms of trigonometric ratios can be transformed into the product forms and the product forms can be transformed into the sum or difference forms. (a) Transformation of product into sum or difference form formula of compound angles; we have sin (A + B) = sinA cosB + cosA. sinB ................ (i) sin (A - B) = sinA. cosB - cosA.sinB ................. (ii) cos (A + B) =cosA.cosB - sinA.sinB ................... (iii) cos (A -B) = cosA. cosB - sinA.sinB ................... (iv) Adding identities (i) and (ii), we get, sin (A + B) + sin (A - B) = 2cosA.sinB. Again, adding identities (iii) and (iv) cos (A + B) + cos(A - B) = 2cosA.cosB Subtracting identity (iii) from(iv), we get, cos(A - B) - cos(A + B) = 2 sinA. sinB Hence, we have the following formulae: (I) 2sinA. cosB = sin(A + B) + sin(A - B) (II) 2cosA. sinB = sin(A + B) - sin(A - B) (III) 2cosA. cosB = cos(A + B) + cos(A - B) (IV) 2sinA.sinB = cos(A -B) – cos(A + B) (b) Transformation of Sum or difference into product Let, A + B = C ..................... (v) and A - B = D ...................... (vi) Adding (v) and (vi), we get, 2A = C + D or, A= C+D 2 Again subtraction (vi) from (v), we get, 2B = C - D or, B= C–D 2 Now, express above four formulae (i), (ii), (iii) and (iv) in terms of C and D. We get, (V) sinC + sinD = 2 sin C + D. cos C-D 2 2 C + D C-D (VI) sinC - sinD = 2cos 2 . sin 2 (VII) cosC + cosD = 2cos C + D . cos C-D 2 2 224 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur