# If sin [cot-

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

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