JEE Main Maths Trig Previous Year Questions With Sol.

Question 1: The general solution of sin x − 3 sin2x + sin3x = cos x − 3 cos2x + cos3x is _________.

Solution:

sinx − 3 sin2x + sin3x = cosx − 3 cos2x + cos3x

⇒ 2 sin2x cosx − 3 sin2x − 2 cos2x cosx + 3 cos2x = 0

⇒ sin2x (2cosx − 3) − cos2x (2 cosx − 3) = 0

⇒ (sin2x − cos2x) (2 cosx − 3) = 0

⇒ sin2x = cos2x

⇒ tan 2x = 1

2x = nπ + (π / 4 )

x = nπ / 2 + π / 8

Question 2: If sec 4θ − sec 2θ = 2, then the general value of θ is __________.

Solution:

sec 4θ − sec 2θ = 2 ⇒ cos 2θ − cos 4θ = 2 cos 4θ cos 2θ

⇒ −cos 4θ = cos 6θ

⇒ 2 cos 5θ cos θ = 0

When cos 5θ = 0, 5θ = (2n + 1)π/2

So θ  = nπ/5 + π/10

= (2n + 1)π/10

When cos θ = 0, θ = (2n+1)π/2.

Question 3: If tan (cot x) = cot (tan x), then sin 2x = ___________.

Solution:

tan (cot x) = cot (tan x) ⇒ tan (cot x) = tan (π / 2 − tan x)

cot x = nπ + π / 2 − tanx

⇒ cot x + tan x = nπ + π / 2

1/sin x cos x = nπ + π / 2

1/sin 2x = nπ/2  + π / 4

⇒ sin2x = 2 / [nπ + {π / 2}]

= 4 / {(2n + 1) π}

Question 4: If the solution for θ of cospθ + cosqθ = 0, p > 0, q > 0 are in A.P., then numerically the smallest common difference of A.P. is ___________.

Solution:

Given cospθ = −cosqθ = cos (π + qθ)

pθ = 2nπ ± (π + qθ), n ∈ I

θ = [(2n + 1)π] / [p − q] or [(2n − 1)π] / [p + q], n ∈ I

Both the solutions form an A.P. θ = [(2n + 1)π] / [p − q] gives us an A.P. with common difference 2π / [p − q] and θ = [(2n − 1)π] / [p + q] gives us an A.P. with common difference = 2π / [p + q].

Certainly, {2π / [p + q]} < {∣2π / [p − q]∣}.

Question 5: If α , β are different values of x satisfying a cosx + b sinx = c, then tan ([α + β] / 2) = ______________.

Solution:

a cosx + b sinx = c ⇒ a {[(1 − tan² (x / 2)] / [1 + tan² (x / 2)]} + 2b {[tan (x / 2) / 1 + tan² (x / 2)} = c

⇒ (a + c) * tan² [x / 2] − 2b tan [x / 2] + (c − a) = 0

This equation has roots tan [α / 2] and tan [β / 2].

Therefore, tan [α / 2] + tan [β / 2] = 2b / [a + c] and tan [α / 2] * tan [β / 2]

= [c − a] / [a + c]

Now

tan ((α + β)/2) = {tan [α / 2] + tan [β / 2]} / {1 − tan [α / 2] * tan [β / 2]}

= {[2b] / [a + c]} / {1− ([c − a] / [a + c])}

= b/a

Question 6: In a triangle, the length of the two larger sides are 10 cm and 9 cm, respectively. If the angles of the triangle are in arithmetic progression, then the length of the third side in cm can be _________.

Solution:

We know that in a triangle larger the side, larger the angle.

Since angles ∠A, ∠B and ∠C are in AP.

Hence, ∠B = 60^o cosB = [a² + c² −b²] / [2ac]

⇒ 1 / 2 = [100 + a² − 81] / [20a]

⇒ a² + 19 = 10a

⇒ a² − 10a + 19 = 0

a = 10 ± (√[100 − 76] / [2])

⇒ a = 5 ± √6

Question 7: In triangle ABC, if ∠A = 45∘, ∠B = 75∘, then a + c√2 = __________.

Solution:

∠C = 180^o − 45^o − 75^o = 60^o

a/sin A = b/sin b = c/sin C

a/sin 45 = b/sin 75 = c/sin 60

=> √2a = 2√2b/(√3+1) = 2c/√3

=> a = 2b/(√3+1)

c = √6b/(√3+1)

a+√2c = [2b/(√3+1)] + [√12b/(√3+1)]

Solving, we get

= 2b

Question 8: If cos⁻¹ p + cos⁻¹ q + cos⁻¹ r = π then p² + q² + r² + 2pqr = ________.

Solution:

Given cos⁻¹ p + cos⁻¹ q + cos⁻¹ r = π

cos⁻¹ p + cos⁻¹ q = π – cos⁻¹ r

cos⁻¹ (pq – √(1 – p²) √(1 – q²) = cos^-1 (-r)

(pq – √(1 – p²) √(1 – q²) = -r

(pq + r) = √(1 – p²) √(1 – q²)

squaring

(pq + r)² = (1 – p²) (1 – q²)

p²q² + 2pqr + r² = 1 – p² – q² + p²q²

p² + q² + r² + 2pqr = 1

Question 9: tan [(π / 4) + (1 / 2) * cos⁻¹ (a / b)] + tan [(π / 4) − (1 / 2) cos^{−1 .}(a / b)] = _________.

Solution:

tan [(π / 4) + (1 / 2) * cos⁻¹ (a / b)] + tan [(π / 4) − (1 / 2) cos^{−1 .}(a / b)]

Let (1 / 2) * cos⁻¹ (a / b) = θ

⇒ cos 2θ = a / b

Thus, tan [{π / 4} + θ] + tan [{π / 4} − θ] = [(1 + tanθ) / (1 − tanθ)]+ ([1 − tanθ] / [1 + tanθ])

= [(1 + tanθ)² + (1 − tanθ)²] / [(1 − tan²θ)]

= [1 + tan²θ + 2tanθ + 1 + tan²θ − 2tanθ] / [(1 – tan²θ)]

= 2 (1 + tan²θ) / [(1 – tan²θ)]

= 2 sec²θ cos²θ/(cos²θ – sin²θ)

= 2 /cos2θ

= 2 / [a / b]

= 2b / a

Question 10: The number of real solutions of tan⁻¹ √[x (x + 1)] + sin⁻¹ [√(x² + x + 1)] = π / 2 is ___________.

Solution:

tan⁻¹ √[x (x + 1)] + sin⁻¹ [√(x² + x + 1)] = π / 2

tan⁻¹ √[x (x + 1)] is defined when x (x + 1) ≥ 0 ..(i)

sin⁻¹ [√x² + x + 1] is defined when 0 ≤ x (x + 1) + 1 ≤ 1 or x² + x + 1  ≥ 1 ..(ii)

From (i) and (ii), x (x + 1) = 0 or x = 0 and -1.

Hence, the number of solution is 2.

Question 11: What is the value of sin (cot⁻¹ x)?

Solution:

Let cot⁻¹ x = θ ⇒ x = cotθ

Now cosecθ = √[1 + cot²θ] = √[1 + x²]

Therefore, sinθ = 1 / cosecθ

= 1 / √[1 + x²]

⇒θ = sin⁻¹ {1 / √[1 + x²]}

Hence, sin (cot⁻¹ x) = sin (sin⁻¹ {1 / √[1 + x²]})

= 1 / √[1 + x²]

= (1 + x²)^−1/2

Question 12: sec² (tan⁻¹ 2) + cosec² (cot⁻¹ 3) = _________.

Solution:

Let (tan⁻¹ 2) = α

⇒ tan α = 2 and cot⁻¹ 3 = β

⇒ cot β = 3

sec² (tan⁻¹ 2) + cosec² (cot⁻¹ 3)

= sec² α + cosec²β

= 1 + tan²α + 1 + cot²β

= 2 + (2)² + (3)²

= 15

Question 13: A vertical pole consists of two parts, the lower part being one-third of the whole. At a point in the horizontal plane through the base of the pole and distance 20 meters from it, the upper part of the pole subtends an angle whose tangent is 1 / 2. The possible heights of the pole are _________.

Solution:

Let H be the height of the pole.

tan α = H/60

tan θ = 1/2

tan (a+b) = (tan a + tan b)/(1-tan a tan b)

tan (α + θ) = H/20

=> (tan α + tan θ)/(1 – tan α tan θ)

H/20 = [(H/60) + (½)]/[(1 – H/120)

=> H² – 80H + 1200 = 0

=> (H – 60) (H-20) = 0

=> H = 60 or H = 20

Height of the pole can be 60 m or 20 m.

Question 14: A tower of height b subtends an angle at a point O on the level of the foot of the tower and a distance a from the foot of the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is ___________.

Solution:

Let AB be the tower and PB be the pole .

Let AB = b and PB = p

tan α = b / a

tan 2α = (p+b)/a

2 tan α/(1 – tan² α) = (p+b)/a

=> (2b/a)/(1 – b²/a²) = (p+b)/a

=> 2ab/(a² – b²) = (p+b)/a

⇒ [2ba² − a²b + b³] / [a² − b²] = p

⇒ p = [b * (a² + b²)] / [a² − b²]

Question 15: A balloon is observed simultaneously from three points A, B and C on a straight road directly under it. The angular elevation at B is twice and at C is thrice that of A. If the distance between A and B is 200 metres and the distance between B and C is 100 metres, then the height of balloon is given by _________.

Solution:

x = h cot3α …..(i)

(x + 100) = h cot2α ……(ii)

(x + 300) = h cotα ……(iii)

From (i) and (ii), −100 = h (cot3α − cot2α),

From (ii) and (iii), −200 = h (cot2α − cotα),

100 = h ([sinα] / [sin3α * sin2α]) and 200 = h ([sinα] / [sin2α * sinα]) or

sin3α / sinα = 200 / 100

⇒ sin3α / sinα = 2

⇒ 3 sinα − 4 sin³α − 2 sinα = 0

⇒ 4 sin³α − sinα = 0

⇒ sinα = 0 or

sin2α = 1 / 4 = sin² (π / 6)

⇒ α = π / 6

Hence, h = 200 * sin [π / 3]

= 200* [√3 / 2]

= 100 √3