Answer :

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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