Prove that 4 tan - 1 1 5 - tan - 1 1 70 + tan - 1 1 99 = π 4 .

Given that LHS = 4 tan - 1 1 5 - tan - 1 1 70 + tan - 1 1 99 We know that tan - 1 A - tan - 1 B = tan - 1 A - B 1 + A B , A B > - 1 and 2 tan - 1 A = tan - 1 2 A 1 - A 2 , - 1 < A < 1 Now, 2 2 tan - 1 1 5 - tan - 1 1 70 - tan - 1 1 99 = 2 tan - 1 2 5 1 - 1 25 - tan - 1 1 70 - 1 99 1 + 1 99 · 1 70 = 2 tan - 1 5 12 - tan - 1 29 6930 6931 6930 = tan - 1 10 12 1 - 25 144 - tan - 1 29 6931 = tan - 1 120 119 - tan - 1 29 6931 = tan - 1 120 119 - 29 6931 1 + 120 119 · 29 6931 = tan - 1 828269 828269 = tan - 1 1 = π 4 = RHS

Solve the equation sin - 1 x + sin - 1 2 x = π 3 .

Since RHS > 0 ⇒ LHS > 0 So, x > 0 Let sin - 1 ⁡ x = θ , θ ∈ ( 0 , π 2 ] , as x > 0 . or sin θ = x ∴ cos ⁡ θ = 1 - x 2 And sin - 1 ⁡ 2 x = ϕ or sin ⁡ ϕ = 2 x ∴ cos ϕ = 1 - 4 x 2 So, cos ⁡ ( θ + ϕ ) = cos ⁡ π 3 ⇒ cos ⁡ θ cos ⁡ ϕ - sin ⁡ θ sin ⁡ ϕ = 1 2 ⇒ 1 - x 2 1 - 4 x 2 - x . 2 x = 1 2 ⇒ 28 x 2 = 3 ⇒ x = 3 28 = 1 2 3 7 , x ≠ - 1 2 3 7 , x > 0

Solve the equation sin - 1 5 x + sin - 1 12 x = π 2 .

Let sin - 1 5 x = θ , then sin ⁡ θ = 5 x ⇒ cosθ = 1 x x 2 - 25 And let sin - 1 ⁡ 12 x = ϕ , then sin ϕ = 12 x ⇒ cosϕ = 1 x x 2 - 144 Now, cos ⁡ ( θ + ϕ ) = cos ⁡ π 2 ⇒ cos ⁡ θcos ⁡ ϕ - sin ⁡ θsin ⁡ ϕ = 0 ⇒ 1 x x 2 - 25 1 x x 2 - 144 - 5 x 12 x = 0 or x 2 x 2 - 169 = 0 Either x 2 = 0 ⇒ x = 0 or x 2 = 169 or x = ± 13 Since, only positive values of x are possible, hence x = 13

Solve the equation tan - 1 a x + tan - 1 b x + tan - 1 c x + tan - 1 d x = π 2 .

Given that tan - 1 a x + tan - 1 b x + tan - 1 c x + tan - 1 d x = π 2 We know that tan - 1 A + tan - 1 B = tan - 1 A + B 1 - A B , A B < 1 and A , B > 0 Now, tan - 1 a x + b x 1 - a x . b x + tan - 1 c x + d x 1 - c x . d x = π 2 ⇒ tan - 1 a + b x x 2 - a b = π 2 - tan - 1 c + d x x 2 - c d ⇒ a + b x x 2 - a b = tan cot - 1 c + d x x 2 - c d ⇒ a + b x x 2 - a b = tan tan - 1 x 2 - c d c + d x = x 2 - c d c + d x ⇒ x 4 - x 2 a b + c d + a c + d a + b c + b d + a b c d = 0 Solving the equation, we can get x .

Solve the equation sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a , where a ≠ b .

Given that sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a So, cos - 1 a x - cos - 1 b x = cos - 1 1 b - cos - 1 1 a Let cos - 1 a x = θ or cos θ = a x and sin θ = 1 x x 2 - a 2 cos - 1 b x = ϕ or cos ϕ = a x and sin ϕ = 1 x x 2 - b 2 cos - 1 1 b = α or cos α = 1 b and sin α = 1 b b 2 - 1 cos - 1 1 a = β or cos ϕ = 1 a and sin β = 1 a a 2 - 1 Now, θ - ϕ = α - β or θ + β = α + ϕ ⇒ cos θ + β = cos α + ϕ ⇒ cos θ cos β - sin θ sin β = cos α cos ϕ - sin α sin ϕ ⇒ a x 1 a - 1 x x 2 - a 2 1 a a 2 - 1 = 1 b b x - 1 x x 2 - b 2 1 b b 2 - 1 ⇒ 1 x - 1 a x x 2 - a 2 a 2 - 1 = 1 x - 1 b x x 2 - b 2 b 2 - 1 ⇒ b 2 a 2 x 2 - a 4 - x 2 + a 2 = a 2 b 2 x 2 - b 4 - x 2 + b 2 ⇒ x 2 a 2 - b 2 = a 2 b 2 a 2 - b 2 ⇒ x = ± a b

Solve the equation 2 tan - 1 x = cos - 1 1 - a 2 1 + a 2 - cos - 1 1 - b 2 1 + b 2 .

Given that 2 tan - 1 x = cos - 1 1 - a 2 1 + a 2 - cos - 1 1 - b 2 1 + b 2 Let a = tan α ⇒ tan - 1 a = α Now, cos 2 a = 1 - tan 2 α 1 + tan 2 α = 1 - a 2 1 + a 2 ⇒ 2 α = cos - 1 1 - a 2 1 + a 2 ⇒ 2 tan - 1 α = cos - 1 1 - a 2 1 + a 2 . . . 1 Similarly, 2 tan - 1 b = cos - 1 1 - b 2 1 + b 2 . . . 2 Given, 2 tan - 1 x = cos - 1 1 - a 2 1 + a 2 - cos - 1 1 - b 2 1 + b 2 From Eqs. ( 1 ) and 2 tan - 1 x = 2 tan - 1 a - 2 tan - 1 b ⇒ tan - 1 x = tan - 1 a - b 1 + a b ⇒ x = a - b 1 + a b

E M B I B E

Plane Trigonometry Part 1>Chapter 15 - Inverse Circular Functions>Examples XLIII

S L Loney Solutions for Chapter: Inverse Circular Functions, Exercise 1: Examples XLIII

Author:S L Loney

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Inverse Circular Functions, Exercise 1: Examples XLIII

Attempt the practice questions on Chapter 15: Inverse Circular Functions, Exercise 1: Examples XLIII with hints and solutions to strengthen your understanding. Plane Trigonometry Part 1 solutions are prepared by Experienced Embibe Experts.