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Question Bank Trigonometry cos3 A + sin 3 A cos3 A – sin 3A + 1. Prove that =2 cos A + sinA cos A – sinA cos3 A + sin 3 A cos3 A – sin 3A Solution. L.H.S. = + cos A + sinA cos A – sinA = (cos A + sinA) (cos 2A + sin 2A – cosA sinA) (cos A – sinA) cos 2A + sin 2 A + cosA sinA + (cos A + sinA) (cos A – sinA) = = = ⎡⎣∵ a 3 + b3 = (a + b) (a 2 + b 2 – ab) and a3 – b3 = (a – b) (a 2 + b 2 + ab)⎤⎦ (cos 2 A + sin 2 A – cosAsinA) + (cos 2 A + sin 2 A + cosA sinA) (1 – cosA sinA) + (1 + cosA sinA) ⎡⎣∵ cos 2 A + sin 2 A = 1⎤⎦ = 1 – cosA sinA + 1 + cosA sinA = 2 = R.H.S. Proved. cosA sinA + 2. Prove that = cosA + sinA 1 – tanA 1 – cotA Solution. cosA sinA L.H.S. = + 1 – tanA 1 – cotA cosA sinA = + sinA cosA 1– 1– cosA sinA cosA × cosA sinA × sinA = + cos A – sinA cos A – sinA cos 2 A sin 2 A = – cos A – sinA cos A – sin A cos 2 A – sin 2 A = cosA – sinA (cosA + sinA) (cosA – sinA) = (cosA – sinA) Math Class X 1 Question Bank [∵a2 – b2 = (a + b) (a – b)] = cosA + sinA = R.H.S. Proved. 3. Prove that = sinA sinA =2+ cotA + cosecA cotA – cosecA Solution. sinA cotA + cosecA sinA sinA sinA = = = cosA 1 cosA + 1 cosA + 1 + sinA sinA sin A 1 – cos 2 A ⎡⎣∵ sin 2 θ = 1 – cos 2 θ ⎤⎦ = cos A + 1 (1 + cosA)(1 – cosA) = = 1 – cosA (cosA + 1) sinA R.H.S. = 2 + cotA – cosecA sinA sin 2 A = 2+ = 2+ cosA 1 cosA – 1 – sinA sinA 1 – cos 2 A ⎡⎣∵ sin 2 θ = 1 – cos 2 θ ⎤⎦ = 2+ cosA – 1 (1 + cosA) (1 – cosA) =2+ cosA – 1 (1 + cosA) (1 – cosA) = 2– cosA – 1 L.H.S. = = 2 – (1 + cosA) = 2 – 1 – cos A = L.H.S. Proved. Math Class X 2 Question Bank 4. If sinA + cosA = m and secA + cosecA = n, prove that n (m2 – 1) = 2 m. Solution. We have, m = sinA + cosA m2 = (sinA + cosA)2 ⇒ = sin2A + cos2A + 2sinA cosA = 1 + 2 sin A cosA 2 m – 1 = 1 + 2 sinA cosA – 1 ⇒ = 2 sinA cosA 2 ⇒ n (m – 1) = (secA + cosecA). 2sinA cosA = 2 sinA cosA secA + 2 sinA cosA cosecA = 2 sinA + 2cosA [∵cosA secA = 1 and sinA cosec A = 1] = 2(sinA + cosA) = 2m 2 Hence, n (m – 1) = 2m. Proved. 5. Prove that 1 1 1 1 – = – (sec A + tanA) cosA cosA (secA – tanA) Solution. 1 1 – secA + tanA cosA 1 1 = – 1 sinA cosA + cos A cosA cosA 1 cos 2 A – 1 – sinA = – = 1 + sinA cosA cos A ( 1 + sinA) L.H.S. = 1 – sin 2 A – 1 – sinA sinA(1 + sinA) = = cos A(1 + sinA) cos A (1 + sinA) = – tanA. Math Class X 3 Question Bank 1 1 – cosA secA – tanA 1 1 – = 1 sinA cosA – cosA cosA 1 cosA = – cosA 1 – sinA 1 – sinA – cos 2 A = cos A (1 – sinA) 1 – sinA – 1 + sin 2 A – sinA(1 – sinA) = = cos A(1 – sinA) cos A(1 – sinA) = – tanA. Hence, LHS = RHS. Proved. 6. If x sin3θ + y cos3θ = sinθ cosθ and x sinθ – y cosθ = 0, then prove that x2 + y2 = 1 Solution. We have x sin3θ + y cos3θ = sinθ cosθ ... (i) x sinθ – y cosθ = 0 ... (ii) sinθ y ⇒ …(iii) = x cosθ sin 2 θ cos 2 θ + y. =1 From (i) x. cosθ sinθ sinθ cosθ sinθ + y. cosθ = 1 ⇒ x. cosθ sinθ y x ⇒ x. sinθ + y. . cosθ = 1 [From (iii)] x y ⇒ y sinθ + x cosθ = 1 ⇒ x cosθ + y sinθ = 1 ... (iv) Squaring (ii) and (iv) and adding, we get, (x sinθ – y cosθ)2 + (x cosθ + y sinθ)2 = 0 + 12 ⇒ x sin2θ – y2 cos2θ – 2xy sinθ cosθ + x2 cos2θ + yi2 sin2 θ + 2xy sinθ cosθ = 1 R.H.S. = Math Class X 4 Question Bank ⇒ x2 + ( sin2 θ + cos2 θ) + y2 (cos2 θ + sin2 θ) = 1 ⇒ x2 + y2 = 1. Proved. cosθ cosθ 7. Is + = 2an identity? If not solve for θ, cosecθ + 1 cosecθ – 1 where 0° < θ < 90°. Solution. cosθ cosθ Here, LHS + cosecθ + 1 cosecθ – 1 cosθ cosθ + = 1 1 +1 –1 sin θ sinθ cosθ sinθ cosθ sinθ = + 1 + sinθ 1 – sinθ cosθ sinθ ( 1– sinθ + 1 + sinθ) = 1 – sin 2 θ 2sinθ cosθ = = 2 tanθ cos 2 θ Thus, the given equality becomes 2 tanθ = 2 If the equality holds true for all values of θ, then the equality is an identity. Let us take θ = 30° 2 So, 2 tan θ = 2 tan30° = 3 ⇒ 2 tan θ ≠ 2 for θ = 30° Therefore the equality is not an identity. It is an equation. Now, 2 tan θ = 2 ⇒ tan θ = 1 ⇒ tan θ = tan 45° ⇒ θ = 45°. 8. If tan θ + sec θ = 3, where θ is acute, then prove that 5 sinθ = 4. Solution. We have tanθ + secθ = 3 sinθ 1 ⇒ + =3 cosθ cosθ 1 + sinθ ⇒ =3 cosθ Math Class X 5 Question Bank (1 + sinθ)2 = 9 cos2θ [Squaring both sides] 2 1 + sin θ + 2 sinθ = 9 – 9 sin2θ 10 sin2θ + 2 sinθ – 8 = 0 10 sin2θ + 10 sinθ – 8 sinθ – 8 = 0 10 sinθ(sinθ + 1) – 8 (sinθ+ 1) = 0 (sinθ + 1) (10 sinθ – 8) = 0 8 4 ⇒ sinθ= –1 or sinθ = = 10 5 4 ⇒ sin θ = [Rejecting sinθ= –1, since θis acute] 5 ⇒ 5 sinθ = 4. Proved. 9. Without using trigonometric tables, prove that : tan 10º tan 20º tan 1 30º tan 70º tan 80º = 3 Solution. L.H.S. = tan 10º tan 20º tan 30º tan 70º tan 80º = (tan 10º tan 80º ), (tan 20º tan 70º) tan 30º = tan (90º – 80º) tan 80º. tan (90º – 70º) tan 70º tan 30º = cot 80º tan 80º. cot70º tan 70º tan 30º [∵ tan (90º – θ ) = cot θ ] 1 1 = tan 70º tan 30º . tan80º. tan 80º tan 70º 1 = 1.1. tan 30º = = R.H.S. Proved. 3 10. Prove that sinA cosA = sec (90º – A ) cosec (90º – A) + sin(90º – A) cos(90º – A) Solution. sinA cosA L.H.S. = + sin(90º – A) (cos(90º – A) sinA cosA = + cosA sinA ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ Math Class X 6 Question Bank = [∵ sin (90º – A) = cosA and cos (90º – A) = sinA] sin 2 A + cos 2 A 1 ⎡⎣∵ sin 2 A + cos 2 A = 1⎤⎦ = = sin A cos A sinA cosA = cosecA secA R.H.S = sec ( 90º – A ) cosec (90º – A) = cosecA secA [∵ sec (90º – A) = cosec A and cosec (90º – A) = secA] = L.H.S. Hence, L.H.S. = R.H.S. Proved. 2 11. Prove that 2 2 sin 20º + sin 70º 2 sin (90º – θ) sinθ + 2 cos 20º + cos 70º tanθ + cos(90º – θ) cosθ cotθ Solution. 2 L.H.S.= 2 sin 20º + sin 70º 2 2 cos 20º + cos 70º 2 + sin (90º – θ) sinθ tan θ + cos(90º – θ) cosθ cotθ sin (90º – 70º ) sin 70º cosθ sinθ sinθ cosθ + + cos 2 (90º – 70º) cos 2 70º tan θ cotθ cos 2 70º + sin 2 70º cosθ sinθ sinθ cosθ = + + sinθ cosθ sin 2 70 + cos 2 70º cos θ sin θ = [∵ sin (90º – θ) = cosθ, cos (90º – θ ) = sinθ] 1 + cos 2 θ + sin 2 θ = 1 = 1 + 1 = 2 = R.H.S. Proved. 12. Using the tables, find the values of (i) sin 60º 23′ (ii) cos 21º 56′ (iii) tan 75º 2′ (iv) cot 40º 36′ Solution. From trigonometric tables, we have (i) sin 60º 18′ = 0.8689 2 2 = Mean difference for 5′ ⇒ (ii) sin 60º 23′ (To be added) = 0.8696 cos 251º54′ Mean difference for 2′ Math Class X =7 = 0.9278 = 2 (To be subtracted) 7 Question Bank =2 ⇒ 21º 56′ = 0.9276 (iii) tan 75º = 3.7321 Mean difference for 2′ = 93 (To be added) ⇒ 75º 2′ = 3.7414 (iv) cot 40º 36′ = cot (90º – 49º 24′) = tan 49º 24′ Now, tan 49º 24′ = 1.1667 13. Find θ when (i) sin θ = 0.0990 (ii) cos θ = 0.5536 (iii) tan θ = 5.2010 Solution. (i) From the table, find the angle whose sine is just smaller than 0.0990 sin θ = 0.0990 sin 5º 36′ = 0.0976 Difference = 0.0014 Mean difference 14 corresponds to 5′ ∴ Required angle = (5º 36′ + 5′= 5º 41′. (ii) From the table, find the angle whose cosine is just greater than 0.5536 cos θ = sin 5º 36′ 0.5536 = 0.5548 Difference = 0.0012 Mean difference 12 corresponds to 5′ ∴ Required angle = (56º 18′ + 5′) = 56º 23′. (iii) From the table, find the angle whose tangent is just smaller than 5.2010 tan θ = 5.2010 tan 79º 6′ = 5.1923 Since mean differences are not given corresponding to 79°, therefore required angle = 79°6′. Math Class X 8 Question Bank 14. A boy standing on a vertical cliff in a jungle observes two rest houses in line with him on opposite sides deep in the jungle below. If their angles of depression are 19° and 26° and the distance between them is 222 m, find the height of the cliff. Solution. Let A be the top of the cliff and C and D be the two rest houses. Let AB = h m and BC = x m Then, BD = (222 – x) m h In ΔABC, tan 19° = x h ⇒ 0.3443 = x h = x × 0.3443 ..(i) ⇒ h In ΔABD, tan 26° = 222 – x h ⇒ 0.4877 = 222 – x h = (222 – x) × 0.4877 ..(ii) ⇒ From (i) and (ii), we have (222 – x) × 0.4877 = x × 0.3443 ⇒ x (0.3443 + 0.4877) = 222 × 0.4877 222 × 0.4877 x = ⇒ 0.832 222 × 0.4877 × 0.3443 From (i), we have, h = 0.832 ⇒ log h = log 222 + log 0.4877 + log 0.3443 – log 0.832 – – – = 2.3463 + 1.6881 + 1.5369 – 1.9201 = 2.3463 + (–1 – 1 + 1) + (0.6881 + 0.5369 – 0.9201) = 2.3463 – 1 + 0.3049 = 2.3463 – 0.6951 = 1.6512 h = antilog 1.6512 = 44.79 ⇒ Hence, height of the cliff = 44.80 m. Math Class X 9 Question Bank 13. An aeroplane is flying horizontally 4000 m above the ground and is going away from an observer on the level ground. At a certain instant the observer finds that the angle of elevation of the plane is 45°. After 15 seconds, its elevation from the same point changes to 30°. Find the speed of the aeroplane in km/h. Solution. Let A be the position of the observer, B be the point whose angle of elevation from A is 45°. Let after 15 seconds the position of the plane be C, whose angle of elevation from A be 30°. BD In ΔABD, tan 45° = AD BD ⇒ 1 = AD ⇒ BD = AD ..(i) CE 4000 [∴ AE = AD + DE] = In ΔACE, tan 30° = AE AD + DE 1 4000 = [From (i)] ⇒ BD + DE 3 1 4000 = ⇒ 4000 + DE 3 ⇒ 4000 3 = 4000 + DE ⇒ DE = 4000 3 – 1 ( ) ⇒ DE = 4000 × 0.732 = 2928 ⇒ Distance covered by the aeroplane in 15 seconds = 2928 m 2928 2928 18 × ⇒ Speed of the aeroplane = m/s = km/h 15 5 15 = 702.72 km/h. Math Class X 10 Question Bank 14. At the foot of a mountain, the elevation of its summit is 45°. After ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain. Solution. Let AB be the mountain of height h m and C be its foot. CD = 1000 m, ∠ACB = 45°, ∠DCB = 30° and ∠ADF = 60°. h In ΔACB, tan 45° = CB h ⇒ 1 = CB ⇒ h = CB ..(i) DE In ΔCDE, sin 30° = 1000 1 DE ⇒ = 2 1000 ⇒ DE = 500 ..(ii) CE In ΔCDE, cos 30° = 1000 CE 3 ⇒ = 1000 2 ⇒ CE = 500 3 ..(iii) Now, BE = BC – EC = h – 500 3 [From (i) and (iii)] AF In Δ ADF, tan 60° = DF h – BE h – 500 ⇒ 3= = [∵ DF = BF and BF = DE] = h – 500 3 h – 500 3 ⇒ h – 500 = h 3 – 1500 ⇒ h 3 –1 = 1000 ( ⇒ Math Class X h= ) 1000 = 1369.86 0.73 11 Question Bank Hence, height of the mountain is 1369.86 m. 15. A man is standing on the deck of a ship which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill. Solution. In the figure. A is the deck of the ship and CD is the hill. Let BC = x m and DE = h m. 8 In ΔABC, tan 30° = x 1 8 ⇒ = ⇒ x = 8 3 m. x 3 DE h In ΔADE, tan 60° = = x AE h ⇒ 3 = [AE = BC = x] x ⇒ h = 3x = 3 × 8 3 = 24 cm. Distance of the hill from the ship = 8 3 m, and height of the hill = (h + 8) m = (24 + 8) m = 32 m. 16. A ladder rests against a house on one side of a street. The angle of elevation of the top of the ladder is 60°. The ladder is turned over to rest against a house on the other side of the street and the elevation now becomes 42° 50'. If the ladder is 40 m long, find the breadth of the street. Solution. In the figure, AB and CD are two houses. O is a point on the street, at which one end of the ladder rests. Let OB = x m and OD = y m. x In ΔAOB, cos 60° = 40 1 x ⇒ = ⇒ x = 20 m. 2 40 Math Class X 12 Question Bank y 40 y ⇒ 0.7333 = [From tables] 40 ⇒ y = 0.7333 × 40 = 29.332 Hence, breadth of the street = (x + y) m = (20 + 29.33) m = 49.33 m. 17. A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is α and that of the top of the h tan α . flagstaff is β. Prove that the height of the tower is tanβ – tanα Solution. Let AB be the tower, AC be the flagstaff of height h m and D be the point of observation. AB In ΔABD, tan α = BD ⇒ AB = BD tan α ..(i) BC In ΔCBD, tan β = BD AB + AC ⇒ tan β = BD ⇒ AB + AC = BD tanβ ⇒ BD tan α + h = BD tanβ ⇒ h = BD (tanβ – tanα) h ⇒ BD = tanβ – tanα ∴ Height of the tower = AB = BD tan α [From (i)] h tan α = tanβ – tanα In ΔCOD, cos 42°50' = Math Class X 13 Question Bank