Maths Quiz on Trigonometry for IIT JEE & Engineering Exam

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]

Solution:

tanα = b / a, tan²α = [2 (b / a)] / [1 − (b / a)²] = [p + b] / a

⇒ 2ba / [a² − b²] = [p + b] / a

⇒ [2ba² − a²b + b³] / [a² − b²] = p

⇒ p = [b * (a² + b²)] / [a² − b²]

Solution:

[H / 3] cotα = d and d = 150 cotφ = 60m or

[H / 3d] = tanα and

[H / d] = tanβ

tan (β − α) = 1 / 2

= {[H / d] − [H / 3d]} / {1 + [H² / 3d²]}

⇒ 1 + [H² / 3d²] = 4H / 3d

⇒ H² − 4dH + 3d2 = 0

⇒ H² − 80H + 3 * (400) = 0

⇒ H = 20 or 60m

a) 15

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 + cosec² α

= 1 + tan²α + 1 + cot²α

= 2 + (2)² + (3)²

= 15

d) 2

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 0 ≤ x (x + 1) ≤ 0 ..(ii)

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

Hence, the number of solution is 2.

b) 2b / a

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

= 2 cos2θ

= 2 / [a / b]

= 2b / a

d) 1

Solution:

According to given condition, we put p = q = r = 1 / 2

Then, p² + q² + r² + 2pqr = (1 / 2)² + (1 / 2)² + (1 / 2)² + 2 * [1 / 2] * [1 / 2] * [1 / 2]

= [1 / 4] + [1 / 4] + [1 / 4] + [2 / 8]

=1

a) 2b

Solution:

∠C = 180^o − 45^o − 75^o = 60^o

Therefore, a + c√2 = k (sin A + √2 sinC)

= k (1 / √2 + [√3 / 2] * √2)

= k ([1 + √3] / [√2]) and

k = b / [sin B]

⇒ a + c√2= b / sin75^o ([1 + √3] / [√2])

= 2b

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 + c√2 = 5 ± √6

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