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X MATHS gb 1 AJAY PARMAR GROUP TUITION Chapter-9 : Trigonometry C Side opposite to angle Trigonometric ratios for angle in a right angled triangle: If in ABC, mB = 90 and mA = then A Side adjacent to angle B Side opposite to angle θ = Hypotenuse Side adjacent t o angle θ cos = = Hypotenuse sin = BC AC AB AC Side opposite to angle θ BC = Side adjacent t o angle θ AB Hypotenuse AC = Side opposite to angle θ BC Hypotenuse AC sec = = Side adjacent t o angle θ AB cosec = AB Side adjacent t o angle θ = Side opposite to angle θ BC tan = The trigonometric ratios remain same for a given angle and even if the lengths of the sides of given triangle are multiplied to some common number, the value of these ratios do not change. The short names of sine, cosine, tangent, cosecant, secant and cotangent are sin, cos, tan, cosec, sec and cot respectively. A is generally represented as A. (sin )2, (cos )2, (tan )2, (cosec )2, (sec )2 and (cot )2 are written as sin2, cos2, tan2, cosec2, sec2 and cot2 respectively. Inter – relationships of trigonometric functions: tan = Identities related to inter – relationships of trigonometric ratios: tan cot = 1, cosec sin = 1, sec cos = 1. Values of trigonometric ratios for 0, 30, 45, 60 and 90. sin θ 1 = ; cosθ cot θ cosec = sin 1 ; sin θ 0 0 cos 1 tan 0 cosec Not defined sec 1 cot Not defined sec = 30 1 2 3 2 1 3 2 45 1 2 1 2 1 2 cot = 1 cos θ 60 3 2 1 2 and cot = cos θ 1 = tan θ sin θ 90 1 0 3 Not defined 2 1 3 2 Not defined 2 3 3 1 1 0 3 Trigonometric ratios for complementary angles: For every R, and 0 < < 90 sin (90 – ) = cos cos (90 – ) = sin tan (90 – ) = cot cot (90 – ) = tan cosec (90 – ) = sec sec (90 – ) = cosec c.g.road :: 98792 12357 :: Gurukul X MATHS gb 2 AJAY PARMAR GROUP TUITION Identities: For every R, and 0 < < 90 sin2 + cos2 = 1 ……(i) sin2 = 1 – cos2 and cos2 = 1 – sin2 Dividing (i) by cos2 we get and Dividing (i) by sin2 we get sin 2 θ 2 cos θ cos 2 θ = 2 cos θ sin 2 θ 1 2 2 cos θ sin θ cos 2 θ 2 sin θ = 1 sin 2 θ tan + 1 = sec , 1 + cot = cosec2, 2 2 sec – tan = 1 and cosec2 – cot2 = 1 and sec2 – 1 = tan2 cosec2 – 1 = cot2 Thus we can express each trigonometric ratio in terms of other trigonometric ratios as follows: sin = 1 cos 2 θ and cos = 1 sin 2 θ cosec = 1 cot 2 θ and cot = cos ec 2θ 1 sec = 1 tan 2 θ and tan = sec 2 θ 1 cosec = and sec = tan = and cot = 2 2 1 1 cos θ sin θ 2 1 sin 2 θ 2 1 1 sin 2 θ cos θ 1 cos 2 θ Exercise – 9.1 1. In ABC, mA = 90. If AB = 5, AC = 12 and BC = 13, find sin C, cos C, tan B, cos B and sin B. sin C = cos C = tan B = AB BC AC BC AC AB 5 13 12 = 13 12 = 5 B = 13 C 12 5 cos B = sin B = AB BC AC BC 5 13 12 = 13 = A 2. In ABC, mB = 90. If BC = 3, AC = 5, find all the six trigonometric ratios of A. ABC, mB = 90. BC = 3, AC = 5 C By Pythagoras theorem AB2 + BC2 = AC2 5 3 2 2 2 AB + 3 = 5 AB2 + 9 = 25 A 4 B AB2 = 25 – 9 = 16 AB = 4 sin A = cos A = tan A = BC AC AB AC BC AB 3. If cos A = 4 5 = = = AC BC AC sec A = = AB AB cot A = = BC 3 5 4 5 3 4 cosec A = , find sin A and tan A. AB 4 = AB = 4k and AC = 5k 5 AC By Pythagoras theorem (4k)2 + BC2 = (5k)2 16k2 + BC2 = 25k2 c.g.road 5 3 5 4 4 3 C Let ABC, mB = 90. cos A = = A 5 4 3 B :: 98792 12357 :: Gurukul X MATHS gb 3 AJAY PARMAR GROUP TUITION BC2 = 25k2 – 16k2 = 9k2 BC = 3k sin A = tan A = BC AC BC AB = 3k = 3 = 5k 3k 4k = 5 3 4 4. If cosec = 13 , find tan and cos . 5 Let ABC, mB = 90 and mA = cosec = AC 13 = AC = 13k and BC = 5k 5 BC By Pythagoras theorem AB2 + (5k)2 = (13k)2 AB2 + 25k2 = 169k2 AB2 = 169k2 – 25k2 = 144k2 AB = 12k 12 BC 12k = = AB 5k 5 12 AB 12k cos = = = AC 13k 13 1 If cos B = , find other five 3 C 13 5 A 12 B tan = 5. trigonometric ratios. B Let ABC, mA = 90. cos B = AB BC = 1 AB = k and BC = 3k 3 By Pythagoras theorem AB2 + AC2 = BC2 (k)2 + AC2 = (3k)2 k2 + AC2 = 9k2 AC2 = 9k2 – k2 = 8k2 AC = 2 2 k sin B = AC 2 2 2 2k = = 3k 3 BC sec B = BC AB = 3k =3 k 3 C 1 A 2 2 tan B = cot B = AC AB AB AC = = 2 2k =2 2 k k 2 2k = cosec B = BC AC 3k = 2 2k = 3 2 2 1 2 2 6. In ABC, mA = 90 and if AB : BC = 1 : 2 find sin B, cos C and tan C. Let ABC, mA = 90. AB : BC = 1 : 2 B AB = k and BC = 2k 2 By Pythagoras theorem 1 AB2 + AC2 = BC2 (k)2 + AC2 = (2k)2 C A 3 k2 + AC2 = 4k2 AC2 = 4k2 – k2 = 3k2 AC = 3 k AC AC 3 3 3k 3k = = cos C = = = 2k 2k 2 2 BC BC 5sinθ 2cosθ 4 7. If tan = , find the value of . 3 3sinθ cosθ 5 sin θ 2 cos θ 3sin θ cos θ sin B = c.g.road :: 98792 12357 :: tan C = AB AC = k 3k = 1 3 Gurukul X MATHS gb 4 AJAY PARMAR GROUP TUITION sin θ cos θ 5 2 cos θ cos θ = [Dividing both numerator and denominator by cos 0] sin θ cos θ 3 cos θ cos θ 4 20 6 5 2 26 3 5 tan θ 2 20 6 = = = 3 = = 12 3 4 3 tan θ 1 12 3 9 3 1 3 3 2sinθ 3cosθ 13 8. If sec = , find the value of . 5 5cosθ 4sinθ Let ABC, mB = 90 and mA = sec = AC AB = 13 5 A AC = 13k and AB = 5k 13 By Pythagoras theorem AB2 + BC2 = (13k)2 (5k)2 + BC2 = 169k2 25k2 + BC2 = 169k2 BC2 = 169k2 – 25k2 = 144k2 BC = 12k C 5 B 12 12 5 AB 5k and cos = = = AC 13k 13 13 12 5 48 15 2 3 39 13 13 2 sin θ 3 cos θ 48 15 = = 13 = =– 25 48 5 cos θ 4 sin θ 25 48 23 5 12 5 4 13 13 13 1 9. If sin B = , prove that 3cos B – 4 cos3 B = 0. 2 C sin = BC AC = 12k 13k = Let ABC, mA = 90. OR AC 1 sin B = = 2 BC 2 AC = k and BC = 2k B 3 1 2 B = 30 1 sin B = A cos B = cos 30 = 3 2 By Pythagoras theorem AB2 + AC2 = BC2 AB2 + (k)2 = (2k)2 AB2 + k2 = 4k2 AB2 = 4k2 – k2 = 3k2 AB = 3 k AB BC cos B = = 3k 3 = 2 2k Now 3 3 – 4 3 = 3 3 – 4 3 3 = 3 3 – 3 3 = 0 = RHS 8 2 2 2 2 2 LHS = 3cos B – 4 cos3 B = 3 10. If tan A = 3 , verify that (1) sin2 A + cos2 A = 1, sec2 A – tan2 A = 1 and 1 + cot2 A = cosec2 A. A Let ABC, mB = 90. OR tan A = BC AB = 3 2 tan A = BC = 3 k and AB = k sin A = sin 60 = C By Pythagoras theorem c.g.road 3 A = 60 1 3 B 1 3 ; cos A = cos 60 = ; 2 2 2 cosec A = cosec 60 = :: 98792 12357 :: 3 ; sec A = sec 60 = 2 Gurukul X MATHS gb 5 AJAY PARMAR GROUP TUITION AC2 = AB2 + BC2 and cot A = cot 60 = 1 3 AC = (k) + ( 3 k) AC2 = (k)2 + ( 3 k)2 AC2 = k2 + 3k2 AC2 = 4k2 AC = 2k 2 2 2 sin A = BC AC = sec A = AC AB = 3k 3 = ; 2 2k 2k = 2; k cos A = AB AC = k 2k cot A = AB BC = k = 3 1 4 4 1 ; 2 = 3k = cosec A = 1 3 AC BC = 2k 3k = 2 3 . Now 2 2 2 (1) LHS = sin A + cos A = 3 1 2 2 2 (2) LHS = sec2 A – tan2 A =(2)2 – 3 2 = 3 1 4 = 4 = 1 = RHS 4 = 4 – 3 = 1 = RHS 2 2 1 2 1 4 3 1 1 + = = = = cosec2A = RHS 3 3 3 3 3 (3) LHS =1 + cot2 A = 1 + 11. If cos = 2 2 verify that tan2 – sin2 = tan2 . sin2 . 3 Let ABC, mB = 90 and mA = cos = AB AC = 2 2 AB = 2 2 k and AC = 3k 3 By Pythagoras theorem AB2 + BC2 = AC2 (2 2 k)2 + BC2 = (3k)2 8k2 + BC2 = 9k2 BC2 = 9k2 – 8k2 = k2 BC = k tan = BC AB = k 2 2k = C 3 1 A 1 sin = and 2 2 2 2 2 BC AC = B k 1 = 3 3k 2 1 1 1 1 1 9 8 = = LHS = tan – sin = = …(i) 72 8 9 72 2 2 3 2 2 2 2 1 1 1 1 1 = = RHS = tan sin = …(ii) 8 9 72 2 2 3 2 2 By (i) and (ii) LHS = RHS 12. In ABC, mB = 90. AC + BC = 25 and AB = 5, determine the values of sin A, cos A and tan A. ABC, mB = 90. AC + BC = 25 A Let BC = x AC = 25 – x 25 – x AB2 + BC2 = AC2 5 2 2 2 5 + x = (25 – x) 25 + x2 = 625 – 50x + x2 C B x 50x = 600 x = 12 BC = 12 AC = 25 – 12 = 13 Thus AB = 5, BC = 12 and AC = 13 c.g.road :: 98792 12357 :: Gurukul X MATHS gb 6 AJAY PARMAR sin A = BC AC GROUP TUITION = 12 ; cos A = 13 AB 5 = ; AC 13 tan A = BC = 12 AB 5 13. In ABC, mC = 90 and mA = mB. (1) Is cos A = cos B? (2) Is tan A = tan B? (3) Will the other trigonometric ratios of A and B be equal? B In ABC, mC = 90 and mA = mB. BC = AC = k suppose k 2k By Pythagoras theorem AC2 = AB2 + BC2 = x2 + x2 = 2x2 k A C AC = 2 x (1) cos A = AB = AC (2) tan A = BC AB = x 2x x x 1 = 2 =1 and cos B = BC AC = and tan B = AB BC = x = 2x 1 2 cos A = cos B. x = 1 tan A = tan B. x (3) The other trigonometric ratios will also be same as the length of adjacent sides, opposite sides and hypotenuse of A and B are also same. 14. If 3cot A = 4, examine whether Let ABC, mB = 90 3cot A = 4 cot A = 4 3 1 tan 2 A = cos2 A – sin2 A. 2 1 tan A C 5 AB 4 = 3 BC 3 AB = 4k and BC = 3k A 4 By Pythagoras theorem AC2 = AB2 + BC2 = (4k)2 + (3k)2 = 16k2 + 9k2 = 25k2 AC = 5k tan A = BC AB = 3k 4k = B 3k 4k 3 AB 4 BC 3 , sin A = = = and cos A = = = 5 AC 5 AC 5k 5k 4 2 3 9 16 9 1 1 1 tan 2 A 7 4 16 LHS = = = = 16 = …(i) 2 2 9 16 9 25 1 tan A 3 1 1 16 16 4 2 2 4 3 16 9 16 9 7 RHS = cos2 A – sin2 A = = = = …(ii) 5 5 25 25 25 25 by (i) and (ii) LHS = RHS 15. If pcot = q, examine whether pcot = q cot = p2 q2 psinθ qcosθ = 2 2. psinθ qcosθ p q q p q p2 q2 sin θ cos θ p q p q p2 q2 p q cot θ p sin θ q cos θ sin θ sin θ p p = LHS = = = = = 2 2 = RHS 2 2 p sin θ q cos θ p q cot θ sin θ cos θ q p q p q p q p q p p sin θ sin θ 16. State whether the following are true or false. Justify your answer. 1. sin = sin = 3 for some angle having 2 Side opposite to angle θ Hypotenuse measure . False. In a right angled triangle the hypotenuse is greater than the side opposite to angle . c.g.road :: 98792 12357 :: Gurukul X MATHS gb 7 AJAY PARMAR GROUP TUITION Side opposite to angle θ <1 Hypotenuse 3 3 sin < 1 but > 1 sin = is not possible. 2 2 2 2. cos = , for some angle having measure . 3 Side adjacent t o angle θ cos = Hypotenuse True. In a right angled triangle the side adjacent to angle is less than hypotenuse. Side opposite to angle θ <1 Hypotenuse 2 2 cos < 1 and < 1 cos = is possible. 3 3 5 3. cosec A = , for some measure of angle A. 2 Hypotenuse cosec A = Side opposite to angle A True. In a right angled triangle the hypotenuse is greater than side opposite to angle A. Hypotenuse 5 5 > 1 and > 1 cosec A = is possible Side opposite to angle A 2 2 4. The value of tan A is always less than 1. False. Side opposite to angle A tan A = Side adjacent t o angle A But in a right angled triangle the side opposite to A can be equal, greater than or smaller than the side adjacent to A. tan A can be 1, greater than 1 or smaller than 1. The given statement is false. 5. sec B = sec B = 3 5 for some B. False. Hypotenuse Side adjacent t o angle B But in a right angled triangle the hypotenuse is greater than side adjacent to angle B. 3 3 Hypotenuse > 1 but < 1 sec B = is not possible. Side adjacent t o angle B 5 5 6. cos = 100 for some angle having measure of . False. cos = Side adjacent t o angle θ Hypotenuse In a right angled triangle the side adjacent to angle is less than hypotenuse. Side opposite to angle θ <1 Hypotenuse cos < 1 but 100 > 1 cos = 100 is not possible. c.g.road :: 98792 12357 :: Gurukul X MATHS gb 8 AJAY PARMAR GROUP TUITION Exercise – 9.2 1. Verify: 1. cos 60 = 1 – 2 sin2 30 = 2 cos2 30 – 1 = cos2 30 – sin2 30. cos 60 = 1 2 …(i) 2 1 1 1 1 1 – 2 sin2 30 = 1 – 2 = 1 – 2 = 1 – = 2 2 2 cos 30 – 1 = 2 3 2 2 4 2 2 …(ii) 3 2 3 3 1 – 1 = 2 – 1 = – 1 = = …(iii) 4 2 3 1 cos2 30 – sin2 30 = – 2 2 2 2 2 2 3 1 2 1 3 1 = = …(iv) = 2 4 4 4 4 = by (i), (ii), (iii) and (iv) cos 60 = 1 – 2 sin2 30 = 2 cos2 30 – 1 = cos2 30 – sin2 30. 2. sin 60 = 2 sin 30 cos 30 LHS = sin 60 = 3 2 …(i) 1 2 RHS = 2sin 30 cos 30 = 2 3 3 = 2 2 …(ii) 2 2 By (i) and (ii) LHS = RHS 3. sin 60 = 2tan30 1 tan 2 30 3 LHS = sin 60 = 2 2 tan 30 RHS = = 1 tan 2 30 …(i) 1 2 3 1 1 3 2 2 = 3 = 3 = 3 1 1 1 3 3 2 3 3 3 = = 4 2 3 4 3 …(ii) By (i) and (ii) LHS = RHS 4. cos 60 = 1 tan 2 30 1 tan 2 30 1 LHS = cos 60 = 2 RHS = 1 tan 2 30 1 tan 2 30 = …(i) 1 1 3 2 1 1 3 2 3 1 1 3 = 3 = 2 = 1 = 3 1 1 2 4 1 3 3 1 …(ii) By (i) and (ii) LHS = RHS 5. cos 90 = 4 cos3 30 – 3 cos 30 LHS = cos 90 = 0 …(i) 3 RHS = 4 cos 30 – 3 cos 30 = 3 4 2 2 3 = 4 3 3 3 3 = 3 3 – 3 3 = 0 …(ii) 2 2 2 2 8 – 3 By (i) and (ii) LHS = RHS 2. Evaluate: 1. sin30 tan45 cosec60 sec30 cos60 cot45 sin 30 tan 45 cos ec60 sec 30 cos 60 cot 45 c.g.road :: 98792 12357 :: Gurukul X MATHS gb 9 AJAY PARMAR GROUP TUITION 1 2 1 2 3 = 2 1 1 3 2 3 2 3 4 = 2 3 4 3 2 3 = 3 34 3 34 = 3 3 4 3 3 4 3 3 4 3 3 4 = (3 3 ) 2 2(3 3) (4) (4) 2 (3 3 ) (4) 2 2 = 27 8 3 16 43 8 3 = 11 27 16 2 3 5cos 60 4sec 2 30 tan 2 45 2 2. sin 2 30 cos 2 30 2 2 1 5 16 67 15 64 12 (1) 2 5 4 1 2 2 2 5 cos 60 4 sec 30 tan 45 67 3 2 4 3 12 = = = = 12 = 2 2 2 1 3 4 1 3 2 12 sin 30 cos 30 1 3 4 4 4 4 2 2 2 3. 2sin2 30 cot 30 – 3 cos2 60 sec2 30 2 1 1 = 2 ( 3 ) 3 2 2 2 2 2 3 1 1 4 = 2 ( 3 ) 3 = 1 = 2 4 4 3 3 3 2 2 4. 3cos2 30 + sec2 30 + 2 cos 0 + 3sin 90 – tan2 60 3cos2 30 + sec2 30 + 2 cos 0 + 3sin 90 – tan2 60 2 3 2 2(1) 3(1) ( 3 ) 2 3 2 2 = 3 9 4 3 4 27 16 24 67 = 3 + 2 + 3 – 3 = + 2 = = 4 3 4 12 3 12 3. In ABC, mB = 90, find the measure of the parts of the triangle other than the one which are A given. 1. mC = 45, AB = 5 ? 5 ? mB = 90, mC = 45 mA = 180 – (90 + 45) = 45 mA = 45 45 ? C B Now as mA = mC AB = BC and also AB = 5 BC = 5 Also sin C = sin 45 = 1 2 = AB AC 5 AC AC = 5 2 2. mA = 30, AC = 10 mB = 90, mA = 30 mC = 180 – (90 + 30) = 60 mC = 60 Now sin C = sin 60 = 3 2 = AB AC AB 10 and sin A = sin 30 = and AB = 5 2 1 BC = 10 2 C 10 BC AC ? ? 30 A ? B BC = 5 3. AC = 6 2 , BC = 3 6 mB = 90, AC = 6 2 , BC = 3 6 cos C = BC AC = 3 6 6 2 = 6 2 3 mC = 30 2 mA = 180 – (90 + 30) = 60 c.g.road A ? ? C :: 98792 12357 :: 3 6 B Gurukul X MATHS gb 10 AJAY PARMAR cos A = AB AC GROUP TUITION 1 AB = AB = 3 2 2 6 2 A 4. AB = 4, BC = 4. mB = 90, AB = BC = 4 mA = mC = 45 Also AC2 = AB2 + BC2 = 42 + 42 = 16 + 16 = 32 = 16 2 AC = 4 2 ? 4 ? ? 4 C B 4. In a rectangle ABCD, AB = 20, mBAC = 60, calculate the length of side BC and diagonals AC and BD . mB = 90 in ABC and mA = 60 D C AB AC 20 60 = AC cos A = cos 1 20 = AC = 40 AC 2 ? ? But in rectangle ABCD AC = BD BD = 40 BC AC BC 3 2 40 ? Also sin A = sin 60 = = 60 BC BC = 20 3 40 A B 20 5. If is a measure of an acute angle and cos = sin , find the value of 2tan + sin2 + 1. cos = sin = 45 ( is an acute angle) 2tan2 + sin2 + 1 2 2 1 1 1 7 6 1 = 2tan 45 + sin 45 + 1 = 2(1) + + 1 = 2(1) + + 1 = 3 + = = 2 2 2 2 2 2 2 2 2 1 tan 2α 2tanα 2 6. If is a measure of an acute angle and 3 sin = 2 cos , prove that =1 2 2 1 tan α 2 2 4 sin α = tan = tan2 = 9 cos α 3 3 5 4 94 = = and 1 + tan2 9 9 9 1 tan α 3 sin = 2 cos 1 – tan2 = 1 – 2 1 tan 2 α 2 tan α 2 LHS = = 2 2 1 tan α 4 5 3 13 3 9 4 9 = 94 9 = 13 9 2 2 2 = 1 tan α 2 5 2 2 9 3 3 3 9 9 =1+ 2 = 5 4 9 13 3 13 2 2 2 5 12 25 144 169 25 144 = = = = = 1 = RHS 169 169 169 169 13 13 7. If A = 30 and B = 60, verify that (1) sin(A + B) = sin A cos B + cos A sin B and (2) cos (A + B) = cos A cos B – sin A sin B. A = 30 and B = 60 A + B = 30 + 60 = 90 (1) LHS = sin(A + B) = sin 90 = 1 …(i) RHS = sin A cos B + cos A sin B = sin 30 cos 60 + cos 30 sin 60 = 1 2 1 2 + 4 3 3 1 3 1 3 = + = = = 1 …(ii) 4 4 4 2 2 4 by (i) and (ii) LHS = RHS (2) cos (A + B) = cos A cos B – sin A sin B. c.g.road :: 98792 12357 :: Gurukul X MATHS gb 11 AJAY PARMAR LHS RHS GROUP TUITION = cos (A + B) = cos 90 = 0 …(i) = cos A cos B – sin A sin B = cos 30 cos 60 – sin 30 sin 60 3 3 1 1 – 2 2 2 2 3 3 = – = 0 …(ii) 4 4 = by (i) and (ii) LHS = RHS 8. If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15 and cos 15. sin (A – B) = sin A cos B – cos A sin B cos (A – B) = cos A cos B + sin A sin B Let A = 45 and B = 30 sin (45 – 30) = sin 45 cos 30 – cos 45 sin 30 cos (45 – 30) = cos 45 cos 30 + sin 45 sin 30 sin 15 = sin 15 = sin 15 = sin 15 = sin 15 = 1 2 3 1 1 – 2 2 2 3 2 2 – cos 15 = 1 cos 15 = 2 2 3 1 cos 15 = 2 2 3 1 2 2 2 cos 15 = 2 6 2 4 1 2 3 1 1 + 2 2 2 3 2 2 + 1 2 2 3 1 2 2 3 1 2 2 2 2 6 2 4 cos 15 = 9. State whether the following are true or false. Justify your answer. 1. The value of sin increases as increases from 0 to 90. True. 1 2 sin 0 = 0, sin 30 = , sin 45 = 1 2 , sin 60 = 3 and sin 90 = 1 2 Thus as the value of increases from 0 to 90, the value of sin also increases. 2. sin = cos for all values of . sin = cos is possible only for 45 for an acute angle. Hence sin cos for all values of . False. 3. cos (A + B) = cos A + cos B. Substituting the value of A and B as 45 False. LHS = sin 90 = 0 and RHS = cos 45 + cos 45 = 1 2 + 1 = 2 2 2 Thus LHS RHS. 4. tan A is not defined for A = 90. tan 90 = sin 90 cos 90 = True. 1 which is not defined. 0 5. The value of cot increases as increases from 0 to 90. cot 0 = not defined, cot 30 = 3 , cot 45 = 1, cot 60 = 1 3 False. and cot 90 = 0 Thus as the value of incrases from 0 to 90, the value of cot decreases. c.g.road :: 98792 12357 :: Gurukul X MATHS gb AJAY PARMAR 12 GROUP TUITION Exercise – 9.3 1. Evaluate: 1. cos 18 sin 72 = cos 18 cos 18 = =1 cos(90 72) cos 18 2. tan 48 – cot 42 = tan 48 – tan (90 – 42) = tan 48 – tan 48 = 0 3. cosec 32 – sec 58 = sec(90 – 32) – sec 58 = sec 58 – sec 58 = 0 4. cos 70 + cos 59 . cosec 31 sin 20 cos 70 = + sin (90 – 59) . cosec 31 sin 20 cos 70 = + sin 31 . cosec 31 cos(90 20) cos 70 = + 1 [ sin . cosec = 1] cos 70 =1+1=2 5. sec 70 sin 20 – cos 20 cosec 70 sec 70 sin 20 – cos 20 cosec 70 = sec 70 cos(90 – 20) – cos 20 sec (90 – 70) = sec 70 cos 70 – cos 20 sec 20 =1–1 [ cos . sec = 1] =0 6. cos(40 – ) – sin (50 + ) + cos 2 40 cos 2 50 sin 2 40 sin 2 50 cos 2 40 sin 2 (90 50) = cos[90 – (40 – )] – sin (50 + ) + 2 sin 40 cos 2 (90 50) = cos[90 – 40 + ] – sin (50 + ) + = cos[50 + ] – sin (50 + ) + 1 =0+1=1 7. cos 2 40 sin 2 40 sin 2 40 cos 2 40 cos 70 cos 55 cos ec35 sin 20 tan 5 tan 25 tan 45 tan 65 tan 85 sin(90 70) cos 55 sec(90 35) = sin 20 tan 5 tan 25 1 cot(90 65) cot(90 85) sin 20 cos 55 sec55 = sin 20 tan 5 tan 25 cot 25 cot 5 cos 55 sec55 =1+ (tan 5 cot 5)(tan 25 cot 25) = 1 + 1 [ cos . sec = 1, tan . cot = 1] =2 8. cot 12 . cot 38 . cot 52 . cot 60 . cot 78 = cot 12 . cot 38 . tan (90 – 52) . = cot 12 . cot 38 . tan 38 . 1 3 . 1 =11 c.g.road 3 = tan (90 – 78) 3 tan 12 = (cot 12 . tan 12) (cot 38 . tan 38) . 1 . 1 [But tan . cot = 1] 3 1 3 :: 98792 12357 :: Gurukul X MATHS gb 13 AJAY PARMAR 9. sin18 + 3 (tan 10 tan 30 tan 40 tan 50 tan 80) cos 72 1 sin 18 = + 3 [tan 10 tan 40 cot (90 – sin(90 72) 3 = sin 18 + sin 18 3[ 1 3 50) cot (90 – 80)] (tan 10 cot 10) (tan 40 cot 40)] = 1 + (1) (1) =1+1=2 10. GROUP TUITION [ tan . cot = 1] cos 58 sin 22 cos 38 cos ec52 sin 32 cos 68 tan 18 tan 35 tan 60 tan 72 tan 55 sin(90 58) sin 22 cos 38 sec(90 52) = sin 32 sin(90 68) tan 18 tan 35 3 cot(90 72) cot(90 55) = sin 32 sin 22 cos 38 sec38 [But cos . sec = 1, tan . cot = 1] sin 32 sin 22 3 (tan 18 cot 18)(tan 35 cot 35) =1+1– 1 3 1 =2– 1 3 = 2 3 1 3 3 3 = 6 3 3 2. Prove the following: 1. sin 48 sec 42 + cos 48 cosec 42 = 2 LHS = sin 48 sec 42 + cos 48 cosec 42 = cos (90 – 48) sec 42 + cos 48 sec (90 – 42) = cos 42 sec 42 + cos 48 sec 48 =1+1 [cos . sec = 1] = 2 = RHS 2. sin70 cosec20 – 2cos 70 cosec 20 = 0 cos20 sec70 sin 70 cos ec 20 LHS = – 2cos 70 cosec 20 cos 20 sec 70 cos(90 70) sec(90 20) = – 2cos 70 sec (90 – cos 20 sec 70 cos 20 sec70 = – 2 cos 70 sec 70 cos 20 sec 70 20) = 1 + 1 – 2(1) [cos . sec = 1] = 2 – 2 = 0 = RHS 3. tan(90 A)cotA 2 – cos2 A = 0 cosec A tan(90 A) cot A LHS = – cos2 A cos ec 2 A cot A cot A = – cos2 A 2 cos ec A cot 2 A 2 = – cos A cos ec 2 A cos 2 A = – cos2 A 2 2 sin A cos ec A = cos2A – cos2A = 0 = RHS 4. [sin . cosec = 1] cos(90 A)sin(90 A) = sin2 A tan(90 A) cos(90 A) sin(90 A) LHS = tan(90 A) c.g.road :: 98792 12357 :: Gurukul X MATHS gb AJAY PARMAR 14 GROUP TUITION = sin A cosA cot A sin A cosA = cos A sin A sin A = 1 sin A = sin2 A = RHS 3. Express the following in terms of trigonometric ratios of angles having measure between 0 and 45: 1. sin 85 + cosec 85 = cos (90 – 85) + sec (90 – 85) = cos 5 + sec 5 2. cos 89 + cosec 87 = sin (90 – 89) + sec (90 – 87) = sin 1 + sec 3 3. sec 81 + cosec 54 = cosec (90 – 81) + sec (90 – 54) = cosec 9 + sec 36 4. For ABC, prove that A C B = cot 2 2 1. tan In ABC, A + B + C = 180 A + C = 180 – B B AC = 90 – 2 2 AC B tan = tan 90 2 2 B AC tan = cot 2 2 BC A cos = sin 2 2 2. In ABC, A + B + C = 180 B + C = 180 – B A BC = 90 – 2 2 B C A cos = cos 90 2 2 A BC cos = sin 2 2 5. If A + B = 90 prove that tanA tanB tanA cotB = sec A. sinA sec B A + B = 90 B = 90 – A LHS = tanA tanB tanA cotB sinA sec B = tanA cot(90 A) tanA tan (90 A) sinA cosec(90 A) = tanA cotA tanA tan A sinA cosecA [But tan . cot = 1, sin . cosec = 1] = 1 tan 2 A 1 [But 1 + tan2A = sec2A] = sec 2 A = sec A = RHS c.g.road :: 98792 12357 :: Gurukul X MATHS gb 15 AJAY PARMAR GROUP TUITION 6. If 3 is the measure of an acute angle and sin 3 = cos ( – 26) then find the value of . sin 3 = cos ( – 26) sin 3 = sin [90 – ( – 26)] sin 3 = sin (90 – + 26) sin 3 = sin (116 – ) 3 = 116 – 4 = 116 = 29 7. If 0 < < 90, and sin = cos 30, then obtain the value of 2tan2 – 1. sin = cos 30 sin = sin (90 – 30) sin = sin 60 = 60 2tan2 – 1 = 2( 3 )2 – 1 = 2(3) – 1 = 6 – 1 = 5 8. If tan A = cot B, prove that A + B = 90, where A and B are measure of acute angles. tan A = cot B tan A = tan (90 – B) A = 90 – B A + B = 90 9. If sec 2A = cosec(A – 42), where 2A is the measure of an acute angle, find the value of A. sec 2A = cosec(A – 42) sec 2A = sec[90 – (A – 42)] sec 2A = sec(90 – A + 42) sec 2A = sec(132 – A) 2A = 132 – A 3A = 132 A = 44 10. If 0 < < 90 and sec = cosec 60, find the value of 2cos2 – 1. sec = cosec 60 sec = sec (90 – 60) sec = sec 30 = 30 2cos – 1 = 2 c.g.road 3 2( 2 2 3 3 2 3 1 )2 – 1 = 2 – 1 = – 1 = = 4 2 2 :: 98792 12357 :: 2 Gurukul X MATHS gb AJAY PARMAR 16 GROUP TUITION Exercise – 9 Prove the following using trigonometric identities: (1 – 19) 1. cos2 + 1 =1 1 cot 2θ 1 2 LHS = cos + 1 cot 2 θ = cos2 + sin2 [ = cos2 + 1 cos ecθ 1 cos ec 2θ = sin ] = 1 = RHS 2. 2sin2 + 3sec2 + 5cot2 + 2cos2 – 3tan2 – 5cosec2 = 0 LHS = 2sin2 + 3sec2 + 5cot2 + 2cos2 – 3tan2 – 5cosec2 = 2sin2 + 2cos2 + 3sec2 – 3tan2 – 5cosec2 + 5cot2 =2(sin2 + cos2) + 3(sec2 – tan2) – 5(cosec2 – cot2) =2(1) + 3(1) – 5(1) = 2 + 3 – 5 = 0 = RHS 3. 4. 1 1 = 2 cosec2 1 cosθ 1 cosθ 1 1 2 2 1 1 LHS = = = = 2 = 2cosec2 1 cos θ 1 cos θ 1 cos θ 1 cos θ 1 cos 2 θ sin θ secθ 1 tanθ sinθ = secθ 1 tanθ sinθ 1 sin θ sin θ 1 sinθ tan θ sinθ cos θ = sec θ 1 = RHS LHS = = cos θ = sin θ tan θ sin θ 1 sec θ 1 sin θ sin θ 1 cos θ cos θ 1 sinθ = sec – tan 1 sinθ 5. LHS = 1 sin θ 1 sin θ (1 sin θ)2 (1 sin θ)2 1 sinθ 1 sinθ = = 2 1 sin θ 1 sin θ 1 sin θ cos 2θ secθ tanθ cosecθ cotθ = cosecθ cotθ secθ tanθ sec θ tan θ LHS = cos ecθ cot θ sec θ tan θ secθ tan θ cosecθ cot θ = cosecθ cot θ cosecθ cot θ secθ tan θ = 6. = RHS = 1 sin θ 1 sin θ = = sec + tan = RHS cos θ cos θ cos θ cosecθ cot θ cosec 2θ cot 2 θ secθ tan θ = 1 cos ecθ cot θ secθ tan θ cos ecθ cot θ = = RHS secθ tan θ cotθ cosecθ 1 7. = cosec + cot cotθ cosecθ 1 cot θ cos ecθ 1 LHS = cot θ cos ecθ 1 cos ecθ cot θ (cos ec 2θ cot 2 θ) = [cosec2 – cot2 = 1] cot θ cos ecθ 1 = sec 2θ tan 2 θ c.g.road :: 98792 12357 :: Gurukul X MATHS gb AJAY PARMAR 17 GROUP TUITION cos ecθ cot θ (cos ecθ cot θ)(cos ecθ cot θ) cot θ cos ecθ 1 (cos ecθ cot θ)[1 (cos ecθ cot θ)] = 1 cos ecθ cot θ (cos ecθ cot θ)(1 cos ecθ cot θ) = 1 cos ecθ cot θ = = cosec + cot = RHS 8. (sin + cosec )2 + (cos + sec )2 = 7 + tan2 + cot2 LHS = (sin + cosec )2 + (cos + sec )2 = sin2 + 2 sin cosec + cosec2 + cos2 + 2 cos sec + sec2 = sin2 + cos2 + 2 + 2 + sec2 + cosec2 [sin cosec = 1 and cos sec = 1] 2 2 = 1 + 2 + 2 + 1 + tan + 1 + cot [sec2 = 1 + tan2 and cosec2 = 1 + cot2] 2 2 = 7 + tan + cot = RHS 9. 2sec2 – sec4 – 2 cosec2 + cosec4 = cot4 – tan4 LHS = 2sec2 – sec4 – 2 cosec2 + cosec4 = sec2 (2 – sec2 ) – cosec2 (2 – cosec2 ) = (1 + tan2 )[2 – (1 + tan2 )] – (1 + cot2 )[2 – (1 + cot2 )] = (1 + tan2 )(2 – 1 – tan2 ) – (1 + cot2 )(2 – 1 – cot2 ) = (1 + tan2 )(1 – tan2 ) – (1 + cot2 )(1 – cot2 ) = (1 – tan4 ) – (1 – cot4 ) = 1 – tan4 – 1 + cot4 = cot4 – tan4 = RHS 10. (sin – sec )2 + (cos – cosec )2 = (1 – sec cosec )2 LHS = (sin – sec )2 + (cos – cosec )2 = (sin – sec )2 + (cos – cosec )2 = sin2 – 2 sin sec + sec2 + cos2 – 2 cos cosec + cosec2 = (sin2 + cos2 ) – 2(sin sec – cos cosec) + (sec2 + cosec2 ) sin θ cos θ 1 1 = (1) – 2 + 2 2 cos θ sin θ cos θ sin θ 2 sin θ cos θ + sin θ cos θ = 1 – 2 cos 2 θ sin 2 θ cos θ sin θ 2 2 2 1 1 + 2 2 cos θ sin θ cos θ sin θ = 1 – 2 = 1 – 2 seccosec + sec2 cosec2 = (1 – sec cosec )2 = RHS 11. 2 sinA cosA sinA cosA = 2 sinA cosA sinA cosA sin A cos 2 A sin A cos A sin A cos A LHS = sin A cos A sin A cos A (sin A cos A) 2 (sin A cos A) 2 = (sin A cos A)(sin A cos A) = = = = 2 1 2cos 2 A sin 2 A cos 2 A 2 sin A cos A sin 2 A cos 2 A 2 sin A cos A (sin 2 A cos 2 A) 2 11 = = RHS(1) 2 2 2 sin A cos A sin A cos 2 A 2 2 1 cos 2 A cos 2 A c.g.road = 1 2 cos 2 A = RHS(2) :: 98792 12357 :: Gurukul X MATHS gb 18 AJAY PARMAR GROUP TUITION 12. tanθ cotθ = sec2 – cosec2 = tan2 – cot2 sinθ.cosθ tan θ cot θ LHS = sin θ cos θ sin θ cos θ = cos θ sin θ sin θ cos θ sin 2 θ cos 2 θ = cos θ sin θ sin θ cos θ = = sin 2 θ cos 2 θ sin 2 θ cos 2 θ sin 2 θ cos 2 θ sin 2 θ cos 2 θ sin 2 θ cos 2 θ 1 1 = cos 2 θ sin 2 θ = sec2 – cosec2 = (1 + tan2 ) – (1 + cot2 ) = 1 + tan2 – 1 – cot2 = tan2 – cot2 13. secθ tanθ = 1 – 2sec tan secθ tanθ sec θ tan θ LHS = sec θ tan θ sec θ tan θ sec θ tan θ = sec θ tan θ sec θ tan θ = + 2 tan2 (sec θ tan θ) 2 sec 2 θ tan 2 θ = sec2 – 2sec tan + tan2 = 1 + tan2 – 2sec tan + tan2 = 1 – 2sec tan + 2tan2 = RHS 14. sec 2θ cosec 2θ = tan + cot LHS = sec 2 θ cosec 2θ = tan 2 θ 1 1 cot 2 θ = tan 2 θ 2 cot 2 θ = tan 2 θ 2 cot 2 θ { tancot = 1) = tan 2 θ 2 cot 2 θ = tan + cot = RHS 15. 1 1 1 1 = cosecA cotA sinA sinA cosecA cotA 1 1 LHS = cos ecA cot A sin A cosec 2 A cot 2 A – cosec A cosecA cot A (cosecA cot A)(cosecA cot A) = – cosec A cosecA cot A RHS = 1 1 sin A cosecA cot A cosec 2 A cot 2 A cosecA cot A (cosecA cot A)(cosecA cot A) = cosec A – cosecA cot A = = cosec A – = cosec A + cot A – cosec A = cot A ……(i) = cosec A – cosec A + cot A = cot A ……(ii) c.g.road :: 98792 12357 :: Gurukul X MATHS gb AJAY PARMAR 19 GROUP TUITION by (i) and (ii) 1 1 1 1 = sin A cos ecA cot A cos ecA cot A sin A tanθ cotθ 16. = 1 + tan + cot = 1 + sec cosec 1 cotθ 1 tanθ tan θ cot θ LHS = 1 cot θ 1 tan θ 1 tan θ = tan θ 1 tan θ 1 1 tan θ tan θ 1 = tan θ 1 tan θ(tan θ 1) tan θ = tan 2 θ 1 tan θ 1 tan θ(tan θ 1) = tan 3 θ 1 tan θ(tan θ 1) = (tan θ 1)(tan 2 θ tan θ 1) tan θ(tan θ 1) = tan 2 θ tan θ 1 tan θ = tan 2 θ tan θ 1 tan θ tan θ tan θ = tan + 1 + cot = 1 + tan + cot = RHS(1) sin θ cos θ = 1 + cos θ sin θ sin 2 θ cos 2 θ = 1 + cos θ sin θ 1 = 1 + cos θ sin θ = 1 + seccosec = RHS(2) 17. sin4 – cos4 = sin2 – cos2 = 2sin2 – 1 = 1 – 2 cos2 LHS = sin4 – cos4 = (sin2 – cos2 )(sin2 + cos2 ) = (sin2 – cos2 )(1) = sin2 – cos2 = RHS(1) = sin2 – (1 – sin2 ) and = (1 – cos2) – cos2 = sin2 – 1 + sin2 and = 1 – cos2 – cos2 = 2sin2 – 1 = RHS(2) and = 1 – 2 cos2 = RHS(3) 18. tan2 A – tan2 B = 2 cos 2 B cos 2 A cos 2 Bcos 2 A 2 = sin 2 A sin 2 B cos 2 Bcos 2 A LHS = tan A – tan B = sin 2 A cos 2 A c.g.road sin 2 B cos 2 B :: 98792 12357 :: Gurukul X MATHS gb 20 AJAY PARMAR = = = = GROUP TUITION sin 2 A cos 2 B sin 2 B cos 2 A cos 2 A cos 2 B (1 cos 2 A) cos 2 B (1 cos 2 B) cos 2 A cos 2 A cos 2 B cos 2 B cos 2 A cos 2 B cos 2 A cos 2 A cos 2 B cos 2 A cos 2 B cos 2 B cos 2 A cos 2 B cos 2 A = RHS(1) and = and = and = sin 2 A(1 sin 2 B) sin 2 B(1 sin 2 A) cos 2 A cos 2 B sin 2 A sin 2 A sin 2 B sin 2 B sin 2 A sin 2 B cos 2 A cos 2 B sin 2 A sin 2 B cos 2 Bcos 2 A = RHS(2) 19. 2(sin6 + cos6 ) – 3(sin4 + cos4 ) + 1 = 0 LHS = 2(sin6 + cos6 ) – 3(sin4 + cos4 ) + 1 = 2(sin2 + cos2 )(sin4 – sin2 cos2 + cos4 ) – 3(sin4 + cos4 ) + 1[ a3 + b3 = (a + b)(a2 – ab + b2)] = 2(1)(sin4 – sin2 cos2 + cos4 ) – 3(sin4 + cos4 ) + 1 = 2sin4 – 2sin2 cos2 + 2cos4 – 3sin4 – 3cos4 + 1 = 1 – sin4 – 2sin2 cos2 – cos4 = 1 – (sin4 + 2sin2 cos2 + cos4 ) = 1 – (sin2 + cos2)2 = 1 – (1)2 = 1 – 1 = 0 = RHS 20. If sin + cos = p and sec + cosec = q then show that q(p2 – 1) = 2p LHS = q(p2 – 1) = (sec + cosec )[(sin + cos )2 – 1] [ substituting the values of q and p] = (sec + cosec )[(sin + cos )2 – 1] 1 1 2 2 = [sin + cos + 2sin cos – 1] cos θ sin θ sin θ cos θ = [1 + 2sin cos – 1] cos θ sin θ sin θ cos θ = [2sin cos ] cos θ sin θ = 2(sin + cos ) = 2p = RHS 21. If tan + sin = a and tan – sin = b then prove that a2 – b2 = 4 ab . tan + sin = a and tan – sin = b ab = (tan + sin )( tan – sin ) = tan2 – sin2 = sin 2 θ cos 2 θ – sin2 = sin2 1 cos 2 θ 1 = sin2 (sec2 – 1) = sin2 tan2 ab = sin tan …(i) LHS = a2 – b2 = (tan + sin )2 – (tan – sin )2 [ substituting the values of a and b] = (tan + sin – tan + sin )(tan + sin + tan – sin ) = (2sin )(2tan ) = 4 sin tan = 4 ab = RHS [ (i)] 22. a cos + b sin = p and a sin – b cos = q then prove that a2 + b2 = p2 + q2. c.g.road :: 98792 12357 :: Gurukul X MATHS gb 21 AJAY PARMAR GROUP TUITION a cos + b sin = p and a sin – b cos = q p = a cos + b sin and q = a sin – b cos 2 2 p = (a cos + b sin ) and q2 = (a sin – b cos )2 p2 = a2cos2 + b2 sin2 + 2abcos sin and q2 = a2sin2 + b2 cos2 – 2abcos sin RHS = p2 + q2 = a2cos2 + b2 sin2 + 2abcos sin + a2sin2 + b2 cos2 – 2abcos sin = a2cos2 + a2sin2 + b2 cos2 + b2 sin2 = a2(cos2 + sin2 ) + b2(cos2 + sin2 ) = a2(1) + b2(1) = a2 + b2 = LHS 23. sec + tan = p, then obtain the value of sec , tan and sin in terms of p. We know that sec2 – tan2 = 1 (sec + tan )(sec – tan ) = 1 But sec + tan = p is given p(sec – tan ) = 1 sec – tan = 1 p …(i) adding (i) and (ii) and Subtracting (ii) from (i) 1 p and 2tan = p – 2sec = p2 1 p and 2tan = sec = p2 1 2p 2p and tan = 2sec = p + cos = and sec – tan = 1 p sec + tan = p …(ii) 1 p p 2 1 p p2 1 2p p2 1 24. Evaluate the following: 1. sec38 cosec52 + 2 3 . tan17 . tan 38 . tan 60 . tan 52 . tan 73 – 3(sin2 32 + sin2 58) 2 . sec 38 + tan17 . tan 38 . 3 . cot (90 – 52) . cot (90 – 73) – 3[sin2 32 + cos2 (90 – 58)] sec( 90 52) 3 sec 38 . = + 2 tan17 . tan 38 . cot 38 . cot 17 – 3[sin2 32 + cos2 32] sec 38 = = 1 + 2 . tan17 . cot 38 . tan 38 . cot 17 – 3[1] = 1 + 2 (tan17 . cot 17)(tan 38 . cot 38) – 3 = 1 + 2 (1)(1) – 3 = 1 + 2 – 3 = 3 – 3 = 0 2. cotθ tan(90 θ) cosecθ sec(90 θ) sin2 37 sin2 53 tan10 tan20 tan30 tan70 tan80 ( cot θ cotθ cos ecθ co sec θ) (sin 2 37 cos 2 (90 53) = 1 tan 10 tan 20 cot(90 70) cot(90 80) 3 = ( cot 2 θ cos ec 2θ) (sin 2 37 cos 2 (90 53) 1 tan 10 tan 20 cot 20 cot 10 3 c.g.road :: 98792 12357 :: Gurukul X MATHS gb 22 AJAY PARMAR GROUP TUITION (cos ec 2θ cot 2 θ) (sin 2 37 cos 2 37) 1 (tan 10 cot 10)(tan 20 cot 20) 3 11 2 = = = 2 3 1 1 (1)(1) 3 3 = 25. If sin A + cos A = 2 sin(90 – A), then obtain the value of cot A. sin A + cos A = 2 sin(90 – A) sin A + cos A = 2 cosA sin A cos A 2 cos A sin A sin A sin A 1 + cotA = 2 cot A 2 cot A – cot A = 1 cotA( 2 – 1) = 1 1 cot A = 2 1 1 = 2 1 2 1 2 1 2 1 = ( 2 ) (1) 2 26. If cosec = 2 , then find the value of 2 2 1 = 2 1 = 2 +1 2sin 2θ 3cot 2θ 4tan 2θ cos 2θ cosec = 2 = 45 sin = sin 45 = 1 2 , cot = cot 45 = 1, tan = tan 45 = 1 and cos = 1 2 2 1 1 3(1) 2 2 2 3(1) 2 1 3 2 sin θ 3 cot θ = 2 = = 2 1 1 4 tan 2 θ cos 2 θ 1 4 4(1) 4(1) 2 2 2 2 (1 sin θ) (2 2sin θ) 8 If tan = then evaluate . (2 2cos θ)(1 cos θ) 15 8 15 1 tan = cot = = tan θ 15 8 2 27. 2 = 1 3 4 2 4 8 = = = 7 8 1 7 7 2 2 2 (1 sin 2 θ) 225 2(1 sin θ)(1 sin θ) (1 sin θ)( 2 2 sin θ) cos 2 θ 15 = = = = cot2 = = 2 2 (2 2 cos θ)(1 cos θ) 2(1 cos θ)(1 cos θ) 64 (1 cos θ) sin θ 8 b 28. If cos = cos = a 2 b2 b a2 b2 , 0 < < 90, find the value of sin and tan . cos2 = sin2 = 1 – cos2 = 1 – sin = b2 a b2 2 b2 = a2 b2 a2 b2 b2 a2 b2 = a2 a2 b2 a a2 b2 a Also tan = sin θ cos θ = a2 b2 b = a b a2 b2 29. Select a proper option from (a), (b), (c) or (d) from given option so that the statement becomes correct: c.g.road :: 98792 12357 :: Gurukul X MATHS gb 23 AJAY PARMAR GROUP TUITION 1. If is a measure of an acute angle such that b sin = a cos , then (a) a2 b2 a b 2 a2 b2 a2 b2 (b) 2 (c) ab a b (d) asinθ bcosθ asinθ bcosθ =…. a b ab b sin = a cos sin : cos = a : b Let sin = ak and cos = bk a sin θ b cos θ a sin θ b cos θ = k (a 2 b 2 ) a(ak ) b(bk ) a2 b2 = 2 2 = 2 2 a(ak ) b(bk ) a b k (a b ) 2. Which of the following is correct for some such that 0 < 90? 1 = 1 (c) sec = 0 sec θ 1 For = 0 sec = 1 =1 sec θ (a) 1 >1 sec θ 3. If tan = (a) 1 5 2 3 1 tan = Now = (b) 5 , then cosec 2θ sec 2θ 1 3 (b) (c) 2 3 1 cos θ <1 =…. (d) 3 cot = 5 cos ec 2θ sec 2 θ cos ec 2θ sec 2 θ 1 cot 2 θ 1 tan 2 θ 1 cot θ 1 tan θ 2 2 4. If tan2 = (a) cosec 2θ sec 2θ (d) 7 8 tan2 = 8 7 2 cot θ tan θ 2 2 , then the value of (b) 8 7 = cot 2 θ tan 2 θ 8 7 cot2 = (c) = 1 ( 5 ) 2 5 2 1 2 ( 5 ) 2 5 (1 sin θ) (1 sin θ) is … (1 cos θ) (1 cos θ) 49 64 (d) 49 64 5 2 = 1 5 25 5 = 25 1 35 1 = 24 2 = 3 36 . 7 8 (1 sin θ)(1 sin θ) cos 2 θ 1 sin 2 θ 7 = = = cot2 = 2 2 8 (1 cos θ)(1 cos θ) 1 cos θ sin θ cosθ sinθ 4 5. If cot = , then the value of 3 cosθ sinθ 4 4 1 (a) 7 (b) (c) (d) – 7 3 3 cos θ sin θ 4 1 cot θ 1 43 cos θ sin θ 1 = sin θ sin θ = =3 = = cos θ sin θ 4 cos θ sin θ 7 cot θ 1 1 4 3 sin θ sin θ 3 4 6. If cosec A = and A + B = 90, then sec B = … . 3 1 4 3 7 (a) (b) (c) (d) 4 3 3 3 A + B = 90 B = 90 – A sec B = sec(90 – A) = cosec A = 4 3 7. If is the measure of an acute angle and 3 sin = cos then = … . (a) 30 (b) 45 (c) 60 (d) 90 c.g.road :: 98792 12357 :: Gurukul X MATHS gb 24 AJAY PARMAR 3 sin = cos cos θ sin θ GROUP TUITION = 3 cot = 3 = 30. 8. If tan A = 5 then find the value of (sin A + cos A) sec A is … . 12 7 12 sin A cos A 1 5 17 (sin A + cos A) sec A = (sin A + cos A) = = tan A + A = +1= cos A 12 12 cos A cos A (a) 12 5 (b) 9. If tan = (a) 1 3 7 12 (c) 17 12 (d) – 1 sin θ =…. 1 sin θ 9 3 (c) (d) 4 16 4 then the value of 3 (b) 3 1 sin θ 1 sin θ 1 sin θ = = 1 sin θ 1 sin θ 1 sin θ (1 sin θ) 2 = (1 sin θ) 2 = 1 sin θ cos θ 9 16 5 25 16 But sec2 = 1 + tan2 = 1 + = = sec = 9 9 9 3 5 4 1 sec – tan = – = 3 3 3 2 2 1 sin θ 1 sin θ = = sec – tan cosθ cos θ cosθ 10. In ABC, if mABC = 90, mACB = 45 and AC = 6, then area of ABC = … . (a) 18 (b) 36 (c) 9 (d) 9 2 mB = 90 and mC = 45 mA = 45 AB = BC = x suppose Also By Pythagoras theorem AB2 + BC2 = AC2 x2 + x2 = 62 2x2 = 36 x2 = 18 Now Area of ABC = ½ AB BC = ½ x x = ½ x2 = ½ 18 = 9 11. If cos2 45 – cos2 30 = x cos 45 sin 45 then x = … . (a) 2 (b) 3 2 (c) – cos2 45 – cos2 30 = x cos 45 sin 45 2 (d) 3 4 2 1 3 x 1 1 2 2 2 2 1 2 1 2 3 4 x 2 2 – 3 = 2x 2x = – 1 x = – 1 2 12. If A and B are complementary angles then sin A sec B = … . (a) 1 (b) 0 (c) – 1 (d) 2 A and B are complementary angles A + B = 90 A = 90 – B sin A sec B = sin (90 – B) sec B = cos B sec B = 1 13. The value of tan 20 tan 25 tan 45 tan 65 tan 70 = … . (a) – 1 (b) 1 (c) 0 (d) 3 tan 20 tan 25 tan 45 tan 65 tan 70 = tan 20 tan 25 1 cot(90 – 65) cot(90 – 70) = tan 20 tan 25 cot 25 cot 20 = (tan 20 cot 20)(tan 25 cot 25) = 1 1 = 1 14. If 7 and 2 are measure of acute angles such that sin 7 = cos 2 then 2sin 3 – 3 tan3 = … . (a) 1 (b) 0 (c) – 1 (d) 1 – 3 sin 7 = cos 2 sin 7 = sin (90 – 2) 7 = 90 – 2 9 = 90 = 10 3 = 30 1 2 2sin 3 – 3 tan3 = 2sin 30 – 3 tan 90 = 2( ) – 3 c.g.road :: 98792 12357 :: 1 3 =1–1=0 Gurukul X MATHS gb 25 AJAY PARMAR 15. If A + B = 90, then GROUP TUITION cotAcotB cotAtanB sin 2 B =…. sinAsecB cos 2 A (a) cot2 B (b) tan2 A A + B = 90 B = 90 – A (c) cot2 A (d) – cot2 A cotAcotB cotAtanB sin 2 B sinAsecB cos 2 A = cotAcot(90 A) cotAtan(90 A) sin 2 (90 A) sinAsec(90 A) cos 2 A cotAtanA cotAcotA cos 2 A =1+ cot2 A – 1 = cot2 A sinAcosecA cos 2 A BC 16. For ABC, sin = … . 2 A A (a) sin (b) sin A (c) cos (d) cos A 2 2 = In ABC, A + B + C = 180 B + C = 180 – B BC = 90 2 BC sin 2 A 2 – A = sin 90 A BC sin = cos 2 2 17. sin4θ cos 4θ sin 2θ cos 2θ =…. (a) 1 (b) 2 sin θ cos θ 4 2 4 sin 2θ cos 2 θ (c) 3 = (d) 0 (sin θ cos θ)(sin θ cos θ) 2 2 2 2 sin 2θ cos 2 θ = sin2 + cos2 = 1 18. If 7cos2 + 3sin2 = 4, then cot = … . (a) 7 (b) 7 3 (c) 3 1 (d) 3 7cos + 3sin = 4 7cos2 + 3(1 – cos2 ) = 4 7cos2 + 3 – 3cos2 ) = 4 4cos2 = 1 2 2 cos2 = 1 4 cos = 1 2 = 60 cot = cot 30 = 3 19. If tan 5 tan 4 = 1 then = … . (a) 7 (b) 3 (c) 10 tan 5 tan 4 = 1 tan 5 = 1 tan4θ (d) 9 tan 5 = cot 4 tan 5 = tan (90 – 4) 5 = 90 – 4 9 = 90 = 10 20. If A and B are the measures of acute angles and tan A= (a) 0 tan A = (b) 1 3 1 2 and sin B = (c) 1 2 3 2 3 and sin B= 1 2 then cos(A + B) = … . 1 2 A = 30 and B = 30 cos(A+B) = cos (30 + 30) = cos 60 = c.g.road (d) 1 1 2 :: 98792 12357 :: Gurukul