Question Bank Trigonometry

Question Bank

Trigonometry

1. Prove that

3 3 3 3

cos A + sin A cos A – sin A +

cos A + sinA cos A – sinA = 2

Solution. L.H.S. =

3 3 3 3

cos A + sin A cos A – sin A +

cos A + sinA cos A – sinA

=

=

= _{(cos A + sin A – cosAsinA) + (cos A + sin A + cosA sinA)}2 2 2 2 = (1 – cosA sinA) + (1 + cosA sinA) ⎡_⎣∵ cos A + sin A = 1 2 2 ⎤_⎦ = 1 – cosA sinA + 1 + cosA sinA = 2 = R.H.S. Proved.

2. Prove that cosA + sinA

1 – tanA 1 – cotA= cosA + sinA Solution.

2 2 2 2

(cos A + sinA) (cos A + sin A – cosA sinA) (cos A – sinA) cos A + sin A + cosA sinA +

(cos A + sinA) (cos A – sinA)

3 3 2 2 3 3 2 2 + = ( + ) ( + – ) and – = ( – ) ( + + ) ⎡ ⎤ ⎣∵ a b a b a b ab a b a b a b ab ⎦ 2 2 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 A sin A = –

cos A – sinA cos A – sin A

2 2

cos A – sin A =

cosA – sinA

(cosA + sinA) (cosA – sinA)

=

[∵a2 – b2 = (a + b) (a – b)]

= cosA + sinA = R.H.S. Proved. 3. Prove that = sinA = 2 + sinA

cotA + cosecA cotA – cosecA Solution. 2 2 2 sinA L.H.S. = cotA + cosecA

sinA sinA sinA

= = =

cosA 1 cosA + 1 _{cosA + 1} +

sinA sinA sin A 1 – cos A = sin = 1 – cos cos A + 1 (1 + cosA)(1 – cosA) = (co ⎡ θ θ⎤ ⎣∵ ⎦ 2 2 2 2 = 1 – cosA sA + 1) sinA R.H.S. = 2 + cotA – cosecA sinA sin A = 2 + = 2 + cosA _– 1 _{cosA – 1} sinA sinA 1 – cos A = 2 + sin = 1 – cos cosA – 1 (1 + cosA = 2 + ⎡ θ θ⎤ ⎣∵ ⎦ ) (1 – cosA) cosA – 1 (1 + cosA) (1 – cosA) = 2 – cosA – 1 = 2 – (1 + cosA) = 2 – 1 – cos A = L.H.S. Proved.

4. If sinA + cosA = m and secA + cosecA = n, prove that n (m2 – 1) =

2m.

Solution. We have,

m = sinA + cosA

⇒ m2 = (sinA + cosA)2

= sin2A + cos2A + 2sinA cosA = 1 + 2 sin A cosA

⇒ m2 – 1 = 1 + 2 sinA cosA – 1

= 2 sinA cosA

⇒ n (m2 – 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

Hence, n (m2 – 1) = 2m. Proved.

5. Prove that 1 – 1 = 1 – 1

(sec A + tanA) cosA cosA (secA – tanA)

Solution. 2 2 1 1 L.H.S. = –

secA + tanA cosA

1 1

= –

1 sinA _cosA

+

cos A cosA

cosA 1 cos A – 1 – sinA

= – =

1 + sinA cosA cos A ( 1 + sinA) 1 – sin A – 1 – sinA sinA(1 + sinA)

= =

cos A(1 + sinA) cos A (1 + sinA) = – tanA.

R.H.S. = 1 – 1

cosA secA – tanA

= 1 – 1 1 sinA cosA _– cosA cosA = 1 – cosA cosA 1 – sinA = 2 1 – sinA – cos A cos A (1 – sinA) = 2

1 – sinA – 1 + sin 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)

⇒ sin = cos θ θ y x …(iii) From (i) 2 2 sin cos + = 1 cos sin x. θ y. θ θ θ

⇒ sin sin + cos cos = 1

cos sin

x. θ θ y. θ θ

θ θ

⇒ x. y sin θ + y. x

x y. cosθ = 1 [From (iii)] ⇒ 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

⇒ x2 + ( sin2 θ + cos2 θ) + y2 (cos2 θ + sin2 θ) = 1

⇒ x2 + y2 = 1. Proved.

7. Is cos + cos cosec + 1 cosec – 1

θ θ

θ θ = 2an identity? If not solve for θ, where 0° < θ < 90°.

Solution.

Here, LHS cos + cos cosec + 1 cosec – 1 θ θ θ θ = cos + cos 1 1 + 1 – 1 sin sin θ θ θ θ

= cos sin +cos sin 1 + sin 1 – sin

θ θ θ θ

θ θ

= cos sin ( 1– sin + 1 + sin )₂ 1 – sin

θ θ θ θ

θ = 2sin cos₂ = 2 tan

cos

θ θ

θ 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°

So, 2 tan θ = 2 tan30° = 2

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

⇒ sin + 1 = 3 cos cos θ θ θ ⇒ 1 + sin = 3 cos θ θ

⇒ (1 + sinθ)2 _{= 9 cos}2_{θ [Squaring both sides]} ⇒ 1 + sin2_{θ + 2 sinθ = 9 – 9 sin}2_θ

⇒ 10 sin2_{θ + 2 sinθ – 8 = 0}

⇒ 10 sin2_{θ + 10 sinθ – 8 sinθ – 8 = 0} ⇒ 10 sinθ(sinθ + 1) – 8 (sinθ+ 1) = 0 ⇒ (sinθ + 1) (10 sinθ – 8) = 0

⇒ sinθ= –1 or sinθ = 8 = 4 10 5

⇒ sin θ = 4

5 [Rejecting sinθ= –1, since θis acute]

⇒ 5 sinθ = 4. Proved.

9. Without using trigonometric tables, prove that : tan 10º tan 20º tan 30º tan 70º tan 80º = 1

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 . tan80º. 1

tan 80º tan 70º tan 70º tan 30º = 1.1. tan 30º = 1

3 = R.H.S. Proved.

10. Prove that

sinA + cosA

sin(90º – A) cos(90º – A) = sec (90º – A ) cosec (90º – A)

Solution.

L.H.S. = sinA + cosA sin(90º – A) (cos(90º – A)

= sinA + cosA cosA sinA

= [∵ sin (90º – A) = cosA and cos (90º – A) = sinA] = 2 2 2 2 sin A cos A 1 = sin A cos 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. 11. Prove that

2 2 2

2 2

sin 20º + sin 70º sin (90º – θ) sinθ cos(90º – θ) cosθ

+ +

cos 20º + cos 70º tanθ cotθ = 2

Solution. L.H.S.=

2 2 2

2 2

sin 20º + sin 70º sin (90º – ) sin cos(90º – ) cos

+ +

cos 20º + cos 70º tan cot

θ θ θ θ

θ θ

=

2 2

2 2

sin (90º – 70º ) sin 70º cos sin sin cos

+ +

cos (90º – 70º) cos 70º tan cot

θ θ θ θ

θ θ

=

2 2

2 2

cos 70º sin 70º cos sin sin cos

+ + sin cos sin 70 + cos 70º cos sin + θ θ θ θ θ θ θ θ

=

[

]

1 θ θ

= 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

Mean difference for 5′ = 7 (To be added) ⇒ sin 60º 23′ = 0.8696

(ii) cos 251º54′ = 0.9278

⇒ 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 = 0.5536 sin 5º 36′ = 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′.

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 In ΔABC, tan 19° = h x ⇒ 0.3443 = h x ⇒ h = x× 0.3443 ..(i) In ΔABD, tan 26° = 222 – h x ⇒ 0.4877 = 222 – h 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

⇒ x = 222 × 0.4877

0.832

From (i), we have, h = 222 × 0.4877 × 0.3443

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

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°. In ΔABD, tan 45° = BD AD ⇒ 1 = BD AD ⇒ BD = AD ..(i) In ΔACE, tan 30° =CE = 4000 AE AD + DE [∴ AE = AD + DE] ⇒ 1 3 = 4000 BD + DE [From (i)] ⇒ 1 3 = 4000 4000 + DE ⇒ 4000 3 = 4000 + DE ⇒ DE = 4000

(

)

⇒ Distance covered by the aeroplane in 15 seconds = 2928 m ⇒ Speed of the aeroplane = 2928

15 m/s =

2928 18

×

15 5 km/h

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°. In ΔACB, tan 45° = CB h ⇒ 1 = CB h ⇒ h = CB ..(i) In ΔCDE, sin 30° = DE 1000 ⇒ 1 2 = DE 1000 ⇒ DE = 500 ..(ii) In ΔCDE, cos 30° = CE 1000 ⇒ 3 2 = CE 1000 ⇒ CE = 500 3 ..(iii)

Now, BE = BC – EC = h – 500 3 [From (i) and (iii)] In Δ ADF, tan 60° = AF DF ⇒ 3 = – BE = – 500 – 500 3 – 500 3 h h h h = [∵DF = BF and BF = DE] ⇒ h – 500 = h 3 – 1500 ⇒ h

(

)

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. In ΔABC, tan 30° = 8 x ⇒ 1 3 = 8 x ⇒ x = 8 3m. In ΔADE, tan 60° = DE AE = h x ⇒ 3 = h x [AE = BC = x] ⇒ h = 3 = 3 × 8 3x = 24 cm.

Distance of the hill from the ship = 8 3m, 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. In ΔAOB, cos 60° = 40 x ⇒ 1 2 = 40 x _⇒ x = 20 m.

In ΔCOD, cos 42°50' = 40 y ⇒ 0.7333 = 40 y [From tables] ⇒ 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 flagstaff is β. Prove that the height of the tower is tan

tanβ – tan

h α α .

Solution. Let AB be the tower, AC be the flagstaff of height h m

and D be the point of observation. In ΔABD, tan α = AB BD ⇒ AB = BD tan α ..(i) In ΔCBD, tan β = BC BD ⇒ tan β = AB + AC BD ⇒ AB + AC = BD tanβ ⇒ BD tan α + h = BD tanβ ⇒ h = BD (tanβ – tanα) ⇒ BD = tanβ – tan h α

∴ Height of the tower = AB = BD tan α [From (i)]

= tan

tanβ – tan

h α