If sin [cot^-1 (x + 1)] = cos(tan^-1 x), then find x.

OR

If (tan^-1 x)² + (cot^-1 x)² = 5π²/8 , then find x. [CBSE 2015]

We know that:

Given,

sin [cot^-1 (x + 1)] = cos(tan^-1 x)

⇒

∵ sin(sin^-1x) = x and cos(cos^-1x) = x

∴

⇒

Squaring both sides we get:

1+x² = 1 + (x+1)²

⇒ (x+1)² – x² = 0

⇒ (x+1-x)(x+1+x) = 0

⇒ 2x+1 = 0

∴ x = -1/2

OR

We know that:

∴

Given,

(tan^-1 x)² + (cot^-1 x)² = 5π²/8

⇒

Let, tan^-1x = y

∴ y² + (π/2 – y)² = 5π²/8

⇒

⇒

⇒ 16y² – 8πy – 3π² = 0

⇒ 16y² – 12πy + 4πy – 3π² = 0

⇒ 4y(4y – 3π) + π(4y – 3π) = 0

⇒ (4y – 3π)(4y + π) = 0

∴ y = 3π/4 or y = -π/4

∴ tan^-1x = 3π/4 or tan^-1x = -π/4

Hence,

x = tan(3π/4) = -1 or x = tan(-π/4) = -1

Hence,

x = -1