If sin (cot ^{ - 1}(x + 1)) = cos (tan ^{ - 1 | vedclass

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Question

(A) Given the equation: $\sin (\cot ^{ - 1}(x + 1)) = \cos (	an ^{ - 1}x)$.First,convert $\cot ^{ - 1}(x + 1)$ to $\sin ^{ - 1}$ form. Let $	heta =

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$ - \frac{1}{2}$

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(A) Given the equation: $\sin (\cot ^{ - 1}(x + 1)) = \cos (	an ^{ - 1}x)$.First,convert $\cot ^{ - 1}(x + 1)$ to $\sin ^{ - 1}$ form. Let $	heta = \cot ^{ - 1}(x + 1)$,then $\cot 	heta = x + 1$. Using the identity $\sin 	heta = \frac{1}{\sqrt{1 + \cot ^2 	heta}}$,we get $\sin 	heta = \frac{1}{\sqrt{1 + (x + 1)^2}} = \frac{1}{\sqrt{x^2 + 2x + 2}}$.Next,convert $	an ^{ - 1}x$ to $\cos ^{ - 1}$ form. Let $\phi = 	an ^{ - 1}x$,then $	an \phi = x$. Using the identity $\cos \phi = \frac{1}{\sqrt{1 + 	an ^2 \phi}}$,we get $\cos \phi = \frac{1}{\sqrt{1 + x^2}}$.Equating the two sides: $\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{1 + x^2}}$.Squaring both sides: $x^2 + 2x + 2 = 1 + x^2$.Subtracting $x^2$ from both sides: $2x + 2 = 1$.Solving for $x$: $2x = -1$,which gives $x = -\frac{1}{2}$.

If $\sin (\cot ^{ - 1}(x + 1)) = \cos (\tan ^{ - 1}x)$,then $x =$

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