Prove that cos - 1 - x = π - cos - 1 x , x ∈ - 1 , 1 .

Let cos A = x , x ∈ - 1 , 1 then, cos π - A = - cos A ⇒ cos π - A = - x So, cos - 1 x = A . . . . i And cos - 1 - x = π - A . . . . . ii Adding equations i and ii , cos - 1 x + cos - 1 - x = A + π - A ⇒ cos - 1 x + cos - 1 - x = π ⇒ cos - 1 - x = π - cos - 1 x

Prove that cos - 1 - x = π - cos - 1 x , x ∈ - 1 , 1 .

Let cos A = x , x ∈ - 1 , 1 then, cos π - A = - cos A ⇒ cos π - A = - x So, cos - 1 x = A . . . . i And cos - 1 - x = π - A . . . . . ii Adding equations i and ii , cos - 1 x + cos - 1 - x = A + π - A ⇒ cos - 1 x + cos - 1 - x = π ⇒ cos - 1 - x = π - cos - 1 x

Prove that 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 .

Consider the given equation: 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 L . H . S . = 2 tan - 1 1 2 + tan - 1 1 7 Using 2 tan - 1 x = tan - 1 2 x 1 - x 2 = tan - 1 2 × 1 2 1 - 1 2 2 + tan - 1 1 7 = tan - 1 1 4 - 1 4 + tan - 1 1 7 = tan - 1 4 3 + tan - 1 1 7 Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x , y > 0 and x y < 1 = tan - 1 4 3 + 1 7 1 - 4 3 × 1 7 = tan - 1 28 + 3 21 21 - 4 21 = tan - 1 31 21 17 21 = tan - 1 31 17 = R . H . S . Hence, proved.

Find the principal value of sin - 1 sin 7 π 4 .

We know that sin - 1 sin x = x , x ∈ - π 2 , π 2 Given sin - 1 sin 7 π 4 = sin - 1 sin 2 π - π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 - sin π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 sin - π 4 ⇒ sin - 1 sin 7 π 4 = - π 4

If cos - 1 x a + cos - 1 y b = θ , then prove that x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ .

Given that, cos - 1 x a + cos - 1 y b = θ and To prove: x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ Using cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 1 - y 2 . cos - 1 x a + cos - 1 y b = θ ⇒ cos - 1 x a × y b - 1 - x a 2 1 - y b 2 = θ ⇒ x a × y b - 1 - x a 2 1 - y b 2 = cos θ ⇒ x y a b - 1 - x 2 a 2 1 - y 2 b 2 = cos θ ⇒ x y a b - cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b - cos θ 2 = 1 - x 2 a 2 1 - y 2 b 2 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - y 2 b 2 - x 2 a 2 + x y a b 2 ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = 1 - cos 2 θ ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = sin 2 θ ∵ cos 2 x + sin 2 x = 1 , x ∈ R Hence, proved.

Solve the equation sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a .

Given that, sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a We know that, sec - 1 x = cos - 1 1 x 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 = b 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 β = 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 ⇒ 1 a x x 2 - a 2 a 2 - 1 = 1 b x x 2 - b 2 b 2 - 1 ⇒ b a 2 x 2 - a 4 - x 2 + a 2 = a b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - a 4 - x 2 + a 2 = a 2 b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - b 2 a 4 - b 2 x 2 + b 2 a 2 = a 2 b 2 x 2 - a 2 b 4 - a 2 x 2 + a 2 b 2 ⇒ a 2 x 2 - b 2 x 2 = b 2 a 4 - a 2 b 4 ⇒ x 2 a 2 - b 2 = a 2 b 2 a 2 - b 2 ⇒ x 2 = a 2 b 2 ⇒ x = ± a b

If cos - 1 x a + cos - 1 y b = θ , then prove that x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ .

Given that, cos - 1 x a + cos - 1 y b = θ and To prove: x 2 a 2 - 2 x y a b cos θ + y 2 b 2 = sin 2 θ Using cos - 1 x + cos - 1 y = cos - 1 x y - 1 - x 2 1 - y 2 . cos - 1 x a + cos - 1 y b = θ ⇒ cos - 1 x a × y b - 1 - x a 2 1 - y b 2 = θ ⇒ x a × y b - 1 - x a 2 1 - y b 2 = cos θ ⇒ x y a b - 1 - x 2 a 2 1 - y 2 b 2 = cos θ ⇒ x y a b - cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b - cos θ 2 = 1 - x 2 a 2 1 - y 2 b 2 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - x 2 a 2 1 - y 2 b 2 ⇒ x y a b 2 + cos 2 θ - 2 x y a b cos θ = 1 - y 2 b 2 - x 2 a 2 + x y a b 2 ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = 1 - cos 2 θ ⇒ y 2 b 2 + x 2 a 2 - 2 x y a b cos θ = sin 2 θ ∵ cos 2 x + sin 2 x = 1 , x ∈ R Hence, proved.

Solve the equation sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a .

Given that, sec - 1 x a - sec - 1 x b = sec - 1 b - sec - 1 a We know that, sec - 1 x = cos - 1 1 x 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 = b 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 β = 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 ⇒ 1 a x x 2 - a 2 a 2 - 1 = 1 b x x 2 - b 2 b 2 - 1 ⇒ b a 2 x 2 - a 4 - x 2 + a 2 = a b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - a 4 - x 2 + a 2 = a 2 b 2 x 2 - b 4 - x 2 + b 2 ⇒ b 2 a 2 x 2 - b 2 a 4 - b 2 x 2 + b 2 a 2 = a 2 b 2 x 2 - a 2 b 4 - a 2 x 2 + a 2 b 2 ⇒ a 2 x 2 - b 2 x 2 = b 2 a 4 - a 2 b 4 ⇒ x 2 a 2 - b 2 = a 2 b 2 a 2 - b 2 ⇒ x 2 = a 2 b 2 ⇒ x = ± a b

Find the principal value of sin - 1 sin 7 π 4 .

We know that sin - 1 sin x = x , x ∈ - π 2 , π 2 Given sin - 1 sin 7 π 4 = sin - 1 sin 2 π - π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 - sin π 4 ⇒ sin - 1 sin 7 π 4 = sin - 1 sin - π 4 ⇒ sin - 1 sin 7 π 4 = - π 4

Prove that 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 .

Consider the given equation: 2 tan - 1 1 2 + tan - 1 1 7 = tan - 1 31 17 L . H . S . = 2 tan - 1 1 2 + tan - 1 1 7 Using 2 tan - 1 x = tan - 1 2 x 1 - x 2 = tan - 1 2 × 1 2 1 - 1 2 2 + tan - 1 1 7 = tan - 1 1 4 - 1 4 + tan - 1 1 7 = tan - 1 4 3 + tan - 1 1 7 Using tan - 1 x + tan - 1 y = tan - 1 x + y 1 - x y ; x , y > 0 and x y < 1 = tan - 1 4 3 + 1 7 1 - 4 3 × 1 7 = tan - 1 28 + 3 21 21 - 4 21 = tan - 1 31 21 17 21 = tan - 1 31 17 = R . H . S . Hence, proved.

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

We have, sin - 1 x + sin - 1 2 x = π 3 ⇒ s i n - 1 x 1 - ( 2 x ) 2 + 2 x 1 - x 2 = π 3 ∵ sin - 1 x + sin - 1 y = sin - 1 x 1 - y 2 + y 1 - x 2 ⇒ s i n - 1 x 1 - 4 x 2 + 2 x 1 - x 2 = π 3 ⇒ x 1 - 4 x 2 + 2 x 1 - x 2 = s i n π 3 ⇒ x 1 - 4 x 2 + 2 x 1 - x 2 = 3 2 ⇒ 2 x 1 - 4 x 2 + 4 x 1 - x 2 = 3 ⇒ 2 x 1 - 4 x 2 = 3 - 4 x 1 - x 2 Squaring both sides, ⇒ 4 x 2 1 - 4 x 2 = 3 + 16 x 2 1 - x 2 - 8 3 x 1 - x 2 ⇒ 8 3 x 1 - x 2 = 16 x 2 1 - x 2 + 3 - 4 x 2 ( 1 - 4 x 2 ) ⇒ 8 3 x 1 - x 2 = 16 x 2 - 16 x 4 + 3 - 4 x 2 + 16 x 4 ⇒ 8 3 x 1 - x 2 = 12 x 2 + 3 Again squaring both sides, ⇒ 192 x 2 ( 1 - x 2 ) = 12 x 2 + 3 2 ⇒ 192 x 2 - 192 x 4 = 144 x 4 + 9 + 72 x 2 ⇒ 336 x 4 - 120 x 2 + 9 = 0 ⇒ 112 x 4 - 40 x 2 + 3 = 0 ⇒ 112 x 4 - 28 x 2 - 12 x 2 + 3 = 0 ⇒ 28 x 2 4 x 2 - 1 - 3 4 x 2 - 1 = 0 ⇒ 28 x 2 - 3 4 x 2 - 1 = 0 ⇒ x 2 = 3 28 or x 2 = 1 4 ⇒ x = ± 3 28 , ± 1 2

If cos - 1 x + cos - 1 y = π 2 , then prove that cos - 1 x = sin - 1 y .

From properties of inverse trigonometric functions for complementary angles, sin - 1 y + cos - 1 y = π 2 , x ∈ - 1 , 1 ⇒ sin - 1 y = π 2 - cos - 1 y Given cos - 1 x + cos - 1 y = π 2 ⇒ cos - 1 x = π 2 - cos - 1 y ⇒ cos - 1 x = sin - 1 y Hence, proved.

If cos - 1 + cos - 1 y + cos - 1 z = π , then prove that x 2 + y 2 + z 2 + 2 x y z = 1 .

Given, cos - 1 x + cos - 1 y + cos - 1 z = π cos - 1 x + cos - 1 y = π – cos - 1 z We know that cos - 1 x + cos - 1 y = cos - 1 [ x y – ( 1 – x 2 ) ( 1 – y 2 ) ] and cos - 1 - x = π - cos - 1 x ⇒ cos - 1 [ x y – ( 1 – x 2 ) ( 1 – y 2 ) ] = cos - 1 ( - z ) ⇒ x y – ( 1 – x 2 ) ( 1 – y 2 ) = ( - z ) ⇒ x y + z = ( 1 – x 2 ) ( 1 – y 2 ) Squaring on both sides, ⇒ ( x y + z ) 2 = [ ( 1 – x 2 ) ( 1 – y 2 ) ] 2 ⇒ x 2 y 2 + z 2 + 2 x y z = ( 1 – x 2 ) ( 1 – y 2 ) ⇒ x 2 y 2 + z 2 + 2 x y z = 1 – x 2 – y 2 + x 2 y 2 ⇒ z 2 + 2 x y z = 1 – x 2 – y 2 ⇒ x 2 + y 2 + z 2 + 2 x y z = 1 Hence, proved.

Prove that tan - 1 n n + 1 - tan - 1 2 n + 1 = - π 4 .

We have to prove: tan - 1 n n + 1 - tan - 1 2 n + 1 = 3 π 4 Consider L . H . S . = tan - 1 n n + 1 - tan - 1 2 n + 1 Using tan - 1 x - tan - 1 y = tan - 1 x - y 1 + x y , x y > - 1 = tan - 1 n n + 1 - 2 n + 1 1 + n n + 1 2 n + 1 = tan - 1 n - n + 1 2 n + 1 n + 1 + n 2 n + 1 = tan - 1 n - 2 n 2 + n + 2 n + 1 n + 1 + 2 n 2 + n = tan - 1 - 2 n 2 - 2 n - 1 2 n 2 + 2 n + 1 = tan - 1 - 2 n 2 + 2 n + 1 2 n 2 + 2 n + 1 = tan - 1 - 1 = - tan - 1 1 = - π 4 ⇒ L . H . S . = R . H . S . Hence, proved.