Mathematics-25.Unit-23 Trigonometric Identities and Equation

SYLLABUS Measurements of Angles, Trigonometric functions, their periodicity and graphs, addition and subtraction formulae, formulae involving multiple and sub-multiple angles, general solution of trigonometric equations. MEASUREMENT OF ANGLES There are two systems used for the measurement of angles. Sexagesimal system: Here a right angle is divided into 90 equal parts known as degrees. Each degree is divided into 60 equal parts called minutes and each minute is further divided into 60 equal parts called seconds. 60 seconds (or 60”) = 1 minute (or 1’) 60 minutes (or 60) = 1 degree (or 1°) 90 degrees (or 90°) = 1 right angle Circular Measurement: In this system a unit called ‘Radian’ is defined as follows: One Radian (1c) = i.e. one radian corresponds to the angle subtended by arc of length ‘r’ at the centre of the circle. Since the ratio is independent of the size of a circle it follows that ‘radian’ is a constant quantity. The circumference of a circle is always equal to  times its diameter or 2 times its radius. For a general angle, e.g. AOD = radian(s). Remember that angles at the centre of a circle are in proportion to the arc. i.e. where AOB = 1c = Note:  is a real number whereas c stands for 180°. Remember the relation  radians = 180° = 200g 1 Radian =  a right angle = = 1800  0.3183098862….. = 57.29577950 = 5701744.8 nearly. Illustration 1: (i). Express 45 20 10 in radian system (ii). The interior angles of a polygon are in A.P. and the smallest angle is 120 and common difference 5. Find the number of sides of the polygon. (iii). Reduce 94g2287 to sexagesimal measure. Solution: (i) 0.7C (ii) n = 9 (iii) 8404853.388 Trigonometric Functions In a right angled triangle ABC, CAB = A and BCA = 90° = /2. AC is the base, BC the altitude and AB is the hypotenuse. We refer to the base as the adjacent side and to the altitude as the opposite side. There are six trigonometric ratios, also called trigonometric functions or circular functions. With reference to angle A, the six ratios are: Obviously, . The reciprocals of sine, cosine and tangent are called the cosecant, secant and cotangent of A respectively. We write these as cosec A, sec A, cot A respectively. Important Notes:  Since the hypotenuse is the greatest side in a right angle triangle, sin A and cos A can never be greater than unity and cosec A and sec A can never be less than unity. Hence sin A 1, cos A 1,cosec A 1, sec A1, while tan A and cot A may have any numerical value lying between -  to + .  All the six trigonometric functions have got a very important property in common that is periodicity. Remember that the trigonometrical ratios are real numbers and remain same so long as the angle remains same. SOME BASIC RESULTS  cos2A + sin2A = 1  cos2A = 1 - sin2A or sin2A = 1 - cos2A  1 + tan2A = sec2A  sec2A - tan2A = 1  cot2A + 1 = cosec2A  cosec2A - cot2A = 1   Fundamental inequalities: For 0 < A < , 0 < cosA < < .  It is possible to express trigonometrical ratios in terms of any one of them e.g. TRIGONOMETRIC RATIOS OF ANY ANGLE Consider the system of rectangular co-ordinate axes dividing the plane into four quadrants. A line OP makes angle  with the positive x-axis. The angle  is said to be positive if measured in counter clockwise direction from the positive x-axis and is negative if measured in clockwise direction. The positive values of the trigonometric ratios in the various quadrants are shown, the signs of the other ratios may be derived. Note that xoy = /2, xox' =  , xoy' = 3/2 PiQi is positive if above the x-axis, negative if below the x-axis, OPi is always taken positive. OQi is positive if along positive x-axis, negative if in opposite direction. (Where i = 1, 2, 3, 4 ) Thus depending on signs of OQi and PiQi the various trigonometrical ratios will have different signs. TABLE  equals sin  cos  tan  cot  Sec  cosec  – – sin cos –tan – cot sec –cosec  90° –  cos sin cot tan cosec sec 90° +  cos – sin –cot – tan –cosec sec 180°–  sin – cos – tan – cot – sec cosec 180°+  – sin – cos tan cot – sec –cosec 360°–  – sin cos – tan – cot sec –cosec 360°+  sin cos tan cot sec cosec Note: • Angle  and 90°– are complementary angles,  and 180°– are supplementary angles • sin(n + (–1)n) = sin, n   • cos(2n ± ) = cos, n   • tan(n + ) = tan, n   i.e. sine of general angle of the form n + (–1)n will have same sign as that of sine of angle  and so on. The same is true for the respective reciprocal functions also. BASIC FORMULAE TABLE 1  sin (A + B) = sin A cos B + cos A sin B  sin (A – B) = sin A cos B – cos A sin B  cos (A + B) = cos A cos B – sin A sin B  cos (A – B) = cos A cos B + sin A sin B    sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A.  cos (A + B) cos (A – B) = cos2 A – sin2 B = cos2 B – sin2 A.  sin2A = 2sinA cosA =  cos2A = cos2A - sin2A = 1-2 sin2A = 2cos2A-1 =  tan2A =  sin3A = 3sinA - 4sin3A = 4sin(60° - A) sinAsin(60° + A)  cos3A = 4cos3A - 3cosA = 4cos(60° - A) cosAcos(60°+A)  TABLE 2     (Here notice (B – A)!)  tanA + tanB =  2sinAcosB = sin(A + B) + sin (A - B)  2cosAsinB = sin(A + B) - sin (A - B)  2cosAcosB = cos(A + B) + cos(A - B)  2sinAsinB = cos(A - B) - cos (A + B) Illustration 2: If in a DABC, cos3A + cos3B + cos3C = 3cosA cosB cosC, then prove that the triangle is equilateral. Solution: Given that cos3A + cos3B + cos3C – 3cosA cosB cosC = 0 Þ (cosA + cosB + cosC) (cos2A + cos2B + cos2C – cosAcosB – cosB cosC – cosCcosA) = 0 Þ cos2A + cos2B + cos2C – cosAcosB – cosBcosC – cosCcosA = 0 (as cosA + cosB + cosC = 1 + 4 sinA/2 sinB/2 sinC/2 ¹ 0) Þ (cosA – cosB)2 + (cosB – cosC)2 + (cosC – cosA)2 = 0 Þ cosA = cosB = cosC Þ A = B = C, Q 0 < A, B, C < p. Þ DABC is equilatral. Illustration 3: If , prove that tan2q = 2tan(3q + a). Solution: = k \ = = Again = \ \ tan2q = 2 tan(3q + a). Illustration 4: For any real q , find the maximum value of cos2( cosq) + sin2(sinq) . Solution: The maximum value of cos2( cosq) is 1 and that of sin2( sinq) is sin21, both exists for q = p/2. Hence maximum value is 1+ sin21. Illustration 5: If . Solution: Þ tan2q = 1/11 now, = 3 – 4 sin2q = 3 - 4 Illustration 6: If A, B,C and D are angles of a quadrilateral and sin sin . sin . sin = , prove that A = B = C = D = /2. Solution: Given Since, A + B = 2 - (C + D), the above equation becomes, = 1 = 0. This is quadratic equation in cos which has real roots. 0 4 Now both cos and cos 1 A = B, C = D. Similarly, A = C, B = D A = B = C = D = /2. Illustration 7: If A, B and C are angles of a triangle, prove that Solution: Since A + B + C = = = = as A, B, C are angles of 0 < A, B, C < sin A, sin B, sin C > 0 E 2 + 2 + 2 E 6 Illustration 8: If cos(A + B) sin (C + D) = cos(A – B) sin(C – D), prove that cotA cotB cotC = cotD. Solution: We have cos (A + B) sin(C + D) = cos(A – B) sin(C – D) i.e. or cotA cotB = tanC cotD or cotA cotB cotC = cotD. Illustration 9: Show that Solution: LHS = Let = a = cos a cos 2a cos 3a cos 4a cos 5a = – cos a cos 2a cos 4a cos 8a cos 5a = – cos 20 a cos 21 a cos 22 a cos 23 a cos 5a = – cos 5a = – = . Illustration 10: Prove that cot 7 = Solution: Let q = 7 Þ 2q = 150 Now cot q = = = . Illustration 11: If 2tan2 a tan2 b tan2 g + tan2 a tan2 b + tan2 b tan2 g + tan2 g tan2 a = 1, prove that sin2 a + sin2 b + sin2 g = 1. Solution: ÞWe have, 2tan2a tan2btan2g + tan2atan2b + tan2btan2g + tan2gtan2a = 1 Þ 2 + cot2g + cot2a + cot2b = cot2a cot2b × cot2g Þ cosec2a + cosec2b + cosec2g – 1 = (cosec2a – 1) (cosec2b – 1) (cosec2g – 1) Þ cosec2a + cosec2b + cosec2g – 1 = – 1 + cosec2a + cosec2b + cosec2g – (cosec2a cosec2b + cosec2b cosec2g + cosec2gcosec2a + cosec2a × cosec2b × cosec2g cosec2a cosec2b + cosec2b × cosec2g + cosec2g cosec2a = cosec2a cosec2b × cosec2g Þ sin2a + sin2b + sin2g = 1 IDENTITIES A trigonometric equation is an identity if it is true for all values of the angle or angles involved. A given identity may be established by reducing either side to the other one, or reducing each side to the same expression, or any convenient modification of these. For any angles A, B, C • sin (A + B +C) = sinA cosB cosC + cosA sinB cosC + cosA cosB sinC - sinA sinB sinC • cos (A + B +C) = cosA cosB cosC- cosA sinB sinC - sinA cosB sinC - sinA sinB cosC • ; • If A, B, C are angles of a triangle (or A + B + C = ): • sinA cosB cosC + cosA sinB cosC + cosA cosB sinC = sinA sinB sinC • cosA sinB sinC + sinA cosB sinC + sinA sinB cosC = 1 + cosA cosB cosC • tanA + tanB + tanC = tanA tanB tanC • cotB cotC + cotC cotA + cotA cotB = 1 • • • sin2A + sin2B + sin2C = 4sinA sinB sinC • cos2A + cos2B + cos2C = -1-4cosA cosB cosC • cos2A + cos2B + cos2C = 1 - 2cosA cosB cosC • • Illustration 12: If x + y + y = xyz, Prove that Solution: Let x = tanA, y = tanB, z = tanC tanA + tanB + tanC = tanA. tanB. tanC. A + B + C = tan(2A + 2B) = tan(2 – 2C) or tan(2A + 2B) = -tan2C or tan2A + tan2B + tan2C = tan2A.tan2B.tan2C or or, Illustration 13: If A + B + C = 1800, prove that sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C. Solution: sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = 4 sin A sin B sin C LHS = sin (B + C – A) + sin (C + A – B) + sin (A + B – C) = sin (p – A – A) + sin (p – B – B) + sin (p – C – C) (Q A + B + C = p) = sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C TRIGONOMETRIC SERIES If we have a cosine series in its product form where the angles are in G.P. with common ratio 2 then multiply both numerator and denominator by 2 sin (least angle). Illustration 14: Simplify the product cosA cos2Acos22A …. Cos2n–1A. Solution: cosA cos2A….cos2n–1A =  (sin2Acos2A)cos2A….cos2n–1A = ……cos2n–1A = ……cos2n–1A Continue like this, finally we have = Note: • where  denotes products . • If we have a cosine series or a sine series in its sum form where the angles are in A.P. then multiply both numerator and denominator with 2sin . Illustration 15: Prove that cos + cos + cos = –1/2. Solution: cos + cos + cos = = . Note: • • . Where  denotes summation. Illustration 16: Sum to n–terms of the series sin – sin( ) + sin( ) – sin( )+ … Solution: sin( ) = –sin and sin(2 ) = sin –sin( ) = sin( + ) sin( ) = sin(2 + ) – sin( ) = sin(3 + ) and so on. Using these results, required sum is S=sin +sin( + )+sin(2 + ) +sin(3 + )+… to n terms = sin + sin ( + ) + sin( + 2 ) + sin( + 3 )+... to n terms = TRIGONOMETRIC EQUATIONS An equation involving one or more trigonometrical ratios of unknown angle is called a trigonometric equation e.g. cos2 x – 4 sin x = 1 It is to be noted that a trigonometrical identity is satisfied for every value of the unknown angle where as trigonometric equation is satisfied only for some values (finite or infinite) of unknown angle. e.g. sec2 x – tan2 x = 1 is a trigonometrical identity as it is satisfied for every value of x  R. SOLUTION OF A TRIGONOMETRIC EQUATION A value of the unknown angle which satisfies the given equation is called a solution of the equation e.g. sin  = ½   = /6 . General Solution Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution. We use the following formulae for solving the trigonometric equations: • sin  = 0   = n, • cos  = 0   = (2n + 1) , • tan  = 0   = n, • sin  = sin    = n + (–1)n, where   [–/2, /2] • cos  = cos    = 2n  , where   [ 0, ] • tan  = tan    = n + , where   ( –/2, /2) • sin2  = sin2  , cos2  = cos2 , tan2  = tan2    = n  , • sin  = 1   = (4n + 1) , • cos  = 1   = 2n , • cos  = –1   = (2n + 1), • sin  = sin  and cos  = cos    = 2n + . Note: • Everywhere in this chapter n is taken as an integer, If not stated otherwise. • The general solution should be given unless the solution is required in a specified interval. •  is taken as the principal value of the angle. Numerically least angle is called the principal value. Method for finding principal value Suppose we have to find the principal value of satisfying the equation sin = – . Since sin is negative, will be in 3rd or 4th quadrant. We can approach 3rd or 4th quadrant from two directions. If we take anticlockwise direction the numerical value of the angle will be greater than . If we approach it in clockwise direction the angle will be numerically less than . For principal value, we have to take numerically smallest angle. So for principal value : 1. If the angle is in 1 st or 2nd quadrant we must select anticlockwise direction and if the angle if the angle is in 3rd or 4th quadrant, we must select clockwise direction. 2. Principal value is never numerically greater than . 3. Principal value always lies in the first circle (i.e. in first rotation) On the above criteria will be or . Among these two has the least numerical value. Hence is the principal value of satisfying the equation sin = – . Algorithm to find the principle argument: Step 1: First draw a trigonometric circle and mark the quadrant, in which the angle may lie. Step 2: Select anticlockwise direction for 1st and 2nd quadrants and select clockwise direction for 3rd and 4th quadrants. Step 3: Find the angle in the first rotation. Step 4: Select the numerically least angle among these two values. The angle thus found will be the principal value. Step 5: In case, two angles one with positive sign and the other with negative sign qualify for the numerically least angle, then it is the convention to select the angle with positive sign as principal value. Example 1: If tan = – 1, then will lie in 2nd or 4th quadrant. For 2nd quadrant we will select anticlockwise and for 4th quadrant. we will select clockwise direction. In the first circle two values and are obtained. Among these two, is numerically least angle. Hence principal value is . Example 2: If cos = , then will lie in 1st or 4th quadrant. For 1st quadrant, we will select anticlockwise direction and for 4th quadrant, we will select clockwise direction. In the first circle two values and are thus found. Both and – have the same numerical value. In such case will be selected as principal value. Illustration 17: Solve cot (sinx + 3) = 1. Solution: sinx + 3 = Þ Þ n = 1 Þ sinx = Þ x = or Illustration 18: If sin 5x + sin 3x + sin x = 0, then find the value of x other than zero, lying between 0  x  . Solution: sin 5x + sin 3x + sin x = 0  (sin 5x + sin x) + sin 3x = 0  2 sin 3x cos 2x + sin 3x = 0  sin 3x(2 cos 2x + 1) = 0  sin 3x = 0; cos 2x = –  3x = n, 2x = 2n  The required value of x is . Illustration 19: Find all acute angle  such that cos  cos 2 cos 4 = . Solution: It is given that cos cos2 cos4 =  2sin cos cos2 cos4 =  2sin2 cos2 cos4 =  2sin4 cos4 = sin  sin8 – sin = 0  2sin cos = 0 Either sin = 0    = For n = 0  = 0 which is not a solution.   = n = 1, i.e.  = or cos  = (2n + 1)  = (2n + 1)   = Hence  = . Illustration 20: Solve for x: . Solution:       sin2x =  1  2x = (2n + 1)  x = (2n+1) , n  I OBJECTIVE ASSIGNMENT 1: The general value of  satisfying both and is : (A) 2n (B) 2n + 7/6 (C) n + /4 (D) 2n + /4 Solution: Let us first find out  lying between 0 and 360°. Since   = 210° or 330° and   = 30° or 210° Hence  = 210° or is the value satisfying both. The general value of Hence (B) is the correct answer. 2: 3 cosec20° - sec20° = (A) 1 (B) 2 (C) 3 (D) 4 Solution: Given = = Hence (D) is the correct answer. 3: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = (A) Cot A (B) tan 6A (C) cot 4A (D) None of these Solution: tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = tanA + 2tan2A + 4tan4A + 8 = cot A Hence (A) is the correct answer. 4: The value of sin 12°. sin48°.sin54° = (A) 1/8 (B) 1/6 (C) 1/4 (D) 1/2 Solution: sin 12°. sin48°.sin54° = = = = = Alternative Method Let  = 12° sin 12°. sin48°.sin54° = = Hence (A) is the correct answer. 5: The smallest positive value of x (in degrees) for which tan(x + 100°) = tan(x + 50°) tan x tan(x - 50°) is : (A) 30° (B) 45° (C) 60° (D) 90° Solution: The relation may be written as     cos50°+ 2sin(2x + 50°) cos(2x + 50°) = 0  cos50°+ sin (4x + 100°) = 0  cos50° + cos(4x + 10°) = 0  cos(2x + 30°) cos(2x – 20°) = 0  x = 30°, 55°  The smallest value of x = 30° Hence (A) is the correct answer. 6. The most general value of  satisfying 3 – 2cos –4sin –cos2 + sin2 =0: (A) 2n (B) 2n + /2 (C) 4n (D) 2n + /4 Solution: 3 – 2cos  – 4 sin  – cos 2 + sin 2 = 0  3 – 2cos  – 4 sin  – 1 + 2sin2  + 2sin  cos  = 0  2sin2  – 2cos – 4sin  + 2sin  cos  + 2 = 0  (sin2  – 2sin  + 1) + cos (sin – 1) = 0  (sin  – 1)[sin  – 1 + cos ] = 0 either sin  = 1   = 2n + /2 where n  I or, sin  + cos  =1 cos(  – /4) = cos(/4)   – /4 = 2n  /4   = 2n, 2n + /2 where n  I Hence  = 2n, 2n + /2. Hence (A, B) is the correct answer. 7: If sinq = 3sin(q + 2a), then the value of tan (q + a) + 2tana is: (A) 0 (B) 2 (C) 4 (D) 1 Solution: Given sin q = 3sin (q + 2a) Þ sin (q + a - a) = 3sin (q + a + a) Þ sin (q + a) cosa – cos(q + a) sina = 3sin (q + a) cosa + 3cos (q + a) sina Þ –2sin (q + a) cosa = 4cos (q + a) sina Þ Þ tan(q+a) + 2tana = 0 Hence (A) is the correct answer. 8: The minimum value of 3tan2q + 12 cot2q is: (A) 6 (B) 8 (C) 10 (D) None of these Solution: A.M. ³ G.M Þ (3tan2q +12 cot2q ) ³ 6 Þ 3 tan2 q +12cot2q has minimum value 12. Hence (D) is the correct answer. 9: If A + B + C = then the value of tanA + tanB + tanC is : (A) 3 (B) 2 (C) > 3 (D) > 2 Solution: tan(A + B) = tan( – C) or, = tanC or, tanA + tanB + tanC = tana tanB tanC [since A.M. G.M.] or, tanA tanB tanC or, A B C 27 [cubing both sides] or tanA tanB tanC 3 tanA + tanB + tanC 3 . Hence (A) is the correct answer. 10: Let 0 < A, B < satisfying the equalities 3 A + 2 B = 1 and 3sin2A – 2sin2B = 0. Then A + 2B = : (A) (B) (C) (D) None of these. Solution: From the second equation, we have sin2B = sin2A …(1) and from the first equality 3 A = 1 –2 B = cos2B …(2) Now cos (A + 2B) = cosA. cos2B – sinA . sin2B = 3 cosA . A – . sinA . sin2A = 3cosA. A – 3 A . cosA = 0 A + 2B = or Given that 0 < A < and 0 < B < 0 < A + 2B < + Hence A + 2B = . Hence (C) is the correct answer. 11: If a cos3 q + 3a cos q sin2 q = x and a sin3 q + 3a cos2 q sin q = y, then (x + y)2/3 + (x – y)2/3 = (A) 2a2/3 (B) a2/3 (C) 3a2/3 (D) 2a1/3 Solution: a cos3 q + 3a cos q sin2 q = x a sin3 q + 3a cos2 q sin q = y x + y = a[sin3 q + cos3 q + 3 sin q cos q(sin q + cos q)] = a(sinq + cosq)3 = sin q + cos q ……(1) x – y = a[cos3 q – sin3q + 3 cosq sin2q – 3 cos2 q sin q] = a[cosq – sinq]3 = cos q – sin q ……(2) (sin q + cos q)2 + (cos q – sin q)2 = 2 (sin2 q + cos2 q) = (x + y)2/3 + (x – y)2/3 = 2a2/3. Hence (A) is the correct answer. 12: If , then sin4a = (A) a/2 (B) a (C) a2/3 (D) 2a Solution: Let a = sin 4q Þ = cos 2q + sin 2q and = cos 2q – sin 2q (1 + ) tan a = (1 + ) Þ (1 + cos 2q + sin 2q) tan a = 1 + cos 2q – sin 2q Þ = cot a Þ = cot a Þ Þ tan = tan Þ q = Þ a = sin 4q = sin (p – 4a) = sin 4 a Hence (B) is the correct answer. 13: If cos2 q = and tan2 = tan2/3a, then cos2/3a + sin2/3a = (A) 2a2/3 (B) (C) (D) 2a1/3 Solution: cos2 q = , tan2 = tan2/3 a tan3 = tan a Þ = k sin3 = k sin a ……(1) cos3 = k cos a ……(2) k2/3 sin2/3 a + k2/3 cos a = 1 sin2/3 a + cos2/3 a = Squaring and adding (1) and (2) k2(sin2 a + cos2 a) = sin6 + cos6 = k2 = 1 – sin2 q = 1 – + cos2 q k2 = Þ k = sin2/3 a + cos2/3 a = . Hence (B) is the correct answer. 14: If 3 sin2 a + 2 sin2 b = 1 and 3 sin 2a –2 sin 2b = 0, where a, b are positive acute angles, then a + 2b = (A) (B) (C) (D) Solution: 3 sin2 a + 2 sin2 b = 1 ……(1) 3 sin 2a = 2 sin 2b ……(2) 3 sin2 a = 1 – 2 sin2 b = cos 2b 3 sin a sin a = cos 2b ……(3) from equation (2) 3 . 2 sin a cos a = 2 sin 2b 3 sin a = from equation (3) sin a = cos 2b cos a cos 2b – sin a sin 2b = 0 cos (a + 2b) = 0 Þ a + 2b = . Hence (A) is the correct answer. 15: The value of is : (A) (B) (C) (D) Solution: = = = Hence (A) is the correct answer. 16: The number of solutions of sin3 x cos x + sin2 x cos2 x + sin x cos3 x = 1 in [0, 2p] is (A) 4 (B) 2 (C) 1 (D) 0 Solution: sin x cos x [sin2x + sin x cos x + cos2 x] = 1 Þ sin x cos x + (sin x cos x)2 = 1 sin2 2x + 2 sin 2x –4 = 0 Þ sin 2x = , which is not possible. Hence (D) is the correct answer. 17: The number of solutions of the equation x3 +2x2 +5x + 2cosx = 0 in [0, 2p] is: (A) 0 (B) 1 (C) 2 (D) 3 Solution: Let f(x) = x3 + 2x2 + 5x +2 cosx Þ f¢(x) = 3x2 +4x + 5 – 2 sinx = 3 Now "x ( as -1 £ sinx £ 1) Þ f¢(x) > 0 " x Þ f(x) is an increasing function. Now f(0) = 2 Þ f(x) = 0 has no solution in [ 0, 2p] . Hence (A) is the correct answer. 18: The value of is equal to (A) -1 (B) (C) (D) Solution: . Hence (D) is the correct answer. 19: sinnx= , where n is an odd natural number, then: (A) = 1, = 2n (B) = 1, = n (C) = 0, = n (D) = 0, = -n Solution: sin nx = Im(ein x) = Im ((cosx + i sinx)n) ….. Since n is odd, let n = 2 + 1 sin nx = – + …. = – + + …. = Hence (C) is the correct answer. 20: If tanx = n. tany, n , then maximum value of (x – y) is equal to: (A) (B) (C) (D) Solution: tanx = n tany, cos(x – y) = cosx. cosy + sinx.siny. cos(x – y) = cosx.cosy(1 + tanx.tany) = cosx. cosy (1 + n tan2y) Now, Hence (D) is the correct answer. 21: If 3sin + 5cos = 5, then the value of 5sin – 3cos is equal to (A) 5 (B) 3 (C) 4 (D) none of these Solution: 3sin = 5(1 – cos) = 5  2sin2/2  tan/2 = 3/5 5sin – 3cos = = Hence (B) is the correct answer. 22: In a ABC, if cotA cotB cotC > 0, then the  is (A) acute angled (B) right angled (C) obtuse angled (D) does not exist Solution: Since cotA cotB cotC > 0 cotA, cotB, cotC are positive   is acute angled Hence (A) is the correct answer. 23: If  < 2 < , then equals to (A) –2cos (B) –2sin (C) 2cos (D) 2sin Solution: = = 2 | sin | = 2sin as Hence (D) is the correct answer. 24: If tan = for some non-square natural number n, then sec2 is (A) a rational number (B) an irrational number (C) a positive number (D) none of these Solution: where n is a non-square natural number so 1 – n  0.  sec2 is a rational number. Hence (A) is the correct answer. 25: The minimum value of cos(cosx) is (A) 0 (B) –cos1 (C) cos1 (D) –1 Solution: cos x varies from –1 to 1 for all real x. Thus cos(cosx) varies from cos1 to cos0  minimum value of cos(cosx) is cos1. Hence (C) is the correct answer. 26: If sin x cos y = 1/4 and 3 tan x = 4 tan y, then find the value of sin (x + y). (A) 1/16 (B) 7/16 (C) 5/16 (D) none of these Solution: 3 tan x = 4 tan y  3 sin x cos y = 4 cos x sin y  3/4 = 4 cos x sin y  cos x sin y = 3/16  sin (x + y) = sin x cos y + cos x sin y = . Hence (B) is the correct answer. 27: The maximum value of 4sin2 x + 3cos2x + is (A) (B) (C) 9 (D) 4 Solution: Maximum value of 4sin2x + 3cos2x i.e. sin2x + 3 is 4 and that of sin + cos is = , both attained at x = /2. Hence the given function has maximum value Hence (A) is the correct answer. 28: If  and  are solutions of sin2 x + a sin x + b = 0 as well as that of cos2x + c cos x + d = 0, then sin( + ) is equal to (A) (B) (C) (D) Solution: According to the given condition, sin+sin = –a and cos +cos= -c.    Hence (D) is the correct answer. 29: If sin, sin and cos are in G.P, then roots of the equation x2 + 2x cot + 1 = 0 are always. (A) equal (B) real (C) imaginary (D) greater than 1 Solution: sin, sin, cos are in G.P.  sin2 = sin cos  cos2 = 1 – sin2  0 Now, the discriminant of the given equation is 4cot2 – 4 = 4 cos2  cosec2  0  Roots are always real. Hence (B) is the correct answer. 30: If then S equals (A) (B) (C) (D) Solution: = = = Hence (C) is the correct answer. 31: If in a ABC, C =90°, then the maximum value of sin A sin B is (A) (B) 1 (C) 2 (D) None Solution: sinA sinB = = = =   Maximum value of sinA sinB = Hence (A) is the correct answer. 32: If in a ABC, sin2A + sin2B + sin2C = 2, then the triangle is always (A) isosceles triangle (B) right angled (C) acute angled (D) obtuse angled Solution: sin2 A + sin2 B + sin2C = 2  2 cos A cos B cos C = 0  either A = 90o or B = 90o or C = 90o Hence (B) is the correct answer. 33. Maximum value of the expression 2sinx + 4cosx + 3 is (A) 2 + 3 (B) 2 - 3 (C) + 3 (D) none of these Solution: Maximum value of 2sinx + 4cosx = 2 . Hence the maximum value of 2sinx + 4cosx +3 is Hence (A) is the correct answer. 34: If sin = 3sin( + 2), then the value of tan ( + ) + 2tan is (A) 3 (B) 2 (C) 1 (D) 0 Solution: Given sin  = 3sin ( + 2)  sin ( +   ) = 3sin ( +  + )  sin ( + ) cos – cos( + ) sin =3sin ( + ) cos + 3cos ( + ) sin  –2sin ( + ) cos = 4cos ( + ) sin   tan(+) + 2tan = 0 Hence (D) is the correct answer. 35: If cos  = , then one of the values of tan is (A) tan cot (B) tan cot (C) sin sin (D) none of these Solution: tan2 = = = = = = tan2 cot2 .  tan =  tan cot . Hence (A) is the correct answer. 36. If tan 2. tan  = 1, then  is equal to (A ) (B) (C) (D) None of these. Solution: tan 2 . tan  = 1 . Hence (B) is the correct answer. 37. If  is the root of 25 , then sin 2 is equal to (A) (B) (C) (D) Solution: Since,  is the root of . Hence (B) is the correct answer. 38. The equation k possesses a solution if (A) k > 6 (B) (C) k > 2 (D) None of these. Solution: We have k But , therefore, Now, Hence (B) is the correct answer. 39. The general solution of the equation tan 3x = tan 5x is (A) x = n/2, n  Z (B) x = n, n  Z (C) x = (2n + 1) , n  Z (D) None of these. Solution: We have tan 3x = tan 5x if n is odd, then x = n/2, gives the extraneous solutions. Thus, the solution of the given equation will be given by x = n/2, where n is even say n = 2 m, m  Z. Hence, the required solution is x = m , m  Z. Hence (B) is the correct answer. 40. The equation is solvable if (A) (B) (C) (D) None of these. Solution: We have where for y to be real. Discriminant . . . (1) But , therefore . . . (2) From (1) and (2), . Hence (B) is the correct answer. 41. The set of values of x for which is (A)  (B) /4 (C) (D) Solution: but this value does not satisfy the given equation as and it reduces to indeterminate form. Hence (A) is the correct answer. 42. If , then  is equal to (A) /3 (B) 2/3 (C) /6 (D) 5/8 Solution: or . Hence (C) is the correct answer. 43. The value of the expression is (A) 1/2 (B) 1 (C) 2 (D) None of these. Solution: Given expression is . Hence (B) is the correct answer. 44. If , then  (only principal value) is (A) /3 (B) 2/3 (C) 4/3 (D) 5/3 Solution: . Hence (A) is the correct answer. 45. Number of solutions of in the interval [0, 2] is (A) 2 (B) 4 (C) 0 (D) None of these. Solution: , but  Solution does not exist. Hence (C) is the correct answer. 46. If , then general solution for  is (A) (B) (C) (D) None of these. Solution: . Hence (B) is the correct answer. 47. Number of solutions of 11 sin x = x is (A) 4 (B) 6 (C) 8 (D) None of these. Solution: 11 sin x = x . . . (1) On replacing n by –, we have 11 sin (–x) = –x So for every positive solution, we have negative solution also and x = 0 is satisfying (1), so number of solution will always be odd. Therefore, (d0 is appropriate choice. Hence (D) is the correct answer. 48. If , then x is equal to (A) (B) (C) (D) None of these. Solution: L.H.S. and equality holds for and R.H.S. equality olds if . Thus L.H.S. = R.H.S. for only. Hence (B) is the correct answer. 49. General solution for  if , is (A) (B) (C) (D) None of these. Solution: . . . (1) and (1) may holds true iff and both equal to 1 simultaneously. First common value of  is for which and and since periodicity of is  and periodicity of is 2, therefore, periodicity of is 2. Therefore, general solution is . Hence (A) is the correct answer. 50. If tan  and tan  are the roots of , then value of tan ( + ) is (A) (B) 1 (C) (D) None of these. Solution: are the roots of and . . Hence (C) is the correct answer. 51. Number of solutions of the equation tan x = sec x = 2 cos x lying in the interval [0, 2] is (A) 0 (B) 1 (C) 2 (D) 3 Solution: The given equation can be written as or –1 Hence, the required number of solutions is 2. Hence (C) is the correct answer. 52. If tan m + cot n  = 0, then the general value of  is (A) (B) (C) (D) Solution: The given equation can be written as or . Hence (A) is the correct answer. 53. The general solution of the equation is (A) (B) (C) (D) Solution: Let or and or Using these in the given equation, we get or or . Hence (D) is the correct answer. 54. One solution of the equation is (A) (B) (C) (D) None of these. Solution: The given equation can be written as or or Either sin  = 0 which gives  = n  or which gives Now, Again Thus, one solution of given equation is . Hence (A) is the correct answer. 55. Solve for x and y, the equations: x y + 3x cosy. y = 14 x y + 3x. y siny = 13 (A) y = where 2n < y < 2n + (B) y = where 2n + < y < 2n + (C) both (D) None of these Solution: Clearly, x 0 dividing the equations, we get by componendo and dividenodo, we get or, = 27 = or, = dividing numerator and denominator by cosy, we get or, . siny = , cosy = (when y is in 1st quadrant) and siny = - and cosy = - (when y is in 3rd quadrant) When y is in first quadrant. When y is in 3rd quadrant. Hence y = where 2n < y < 2n + and y = where 2n + < y < 2n + 56. The solution of sinx + cosx = is : (A) 2np + (B) 2np - (C) (D) None of these Solution: Given, cosx + sinx = Þ cos x + sinx = Þ cos Þ Þ x = 2np ± . Þ x = 2np + , 2np - where n Î I. Hence (A, B) is the correct answer. 57. The solution of the equation tan  . tan 2 = 1 is : (A) np + (B) np - (C) (D) n  Solution: Given tan . tan 2 = 1  = 1  2 tan2  = 1 –tan2   3 tan2  = 1  tan  =   = n  Hence (D) is the correct answer. 58. Find the general solution of the equation sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x: (A) (B) np - (C) (D) n  Solution: Given sin x – 3 sin 2x + sin 3x = cos x –3 cos 2x + cos 3x  2 sin 2x cos x – 3 sin 2x = 2 cos x cos 2x – 3 cos 2x  sin 2x (2 cos x –3) = cos 2x (2 cos x –3)  sin 2x = cos 2x ( cos x  3/2)  tan 2x = 1  2x = n +  x = , n  I. Hence (C) is the correct answer. 59. Solve for x, the equation sin3x + sin x cos x + cos3x = 1: (A) 2mp (B) (4n + 1) (C) Both (D) None of these Solution: The given equation is sin3 x + cos3 x + sin x cos x = 1  (sin x + cos x) (sin2 x – sin x cos x + cos2 x ) + sin x cos x – 1 = 0  (1 – sin x cos x)[sin x + cos x – 1] = 0 Either 1 – sin x cos x = 0  sin 2 x = 2 which is not possible Or, sin x + cos x – 1 = 0  cos (x – /4) =    x = 2m and x = (4n + 1) Hence (C) is the correct answer. 60. The equation esinx – e–sinx – 4 = 0 has: (A) no real solution (B) one real solution (C) two real solutions (D) can't be determined Solution: The given equation can be written as e2 sin x – 4esin x – 1 = 0  esin x = = 2 +  sin x = ln (2 + ) (ln (2 – ) not defined as (2 – ) is negative) Now, 2 + > e  ln (2 + ) > 1  sin x > 1 Which is not possible. Hence no real solution. Hence (A) is the correct answer. 61. If tan ( cos x) = cot ( sin x), then is (A) (B) (C) 0 (D) None of these. Solution: Given that tan ( cos x) = cos ( sin x) or . Hence (B) is the correct answer.