The value of x for which sin(cot^{-1} (1 + x) | vedclass

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(D) We are given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(	an^{-1} x)$.First,simplify the right-hand side: Let $	an^{-1} x = 	heta$,then $x = \

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$-\frac{1}{2}$

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(D) We are given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(	an^{-1} x)$.First,simplify the right-hand side: Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$. Since $\cos 	heta = \frac{1}{\sec 	heta} = \frac{1}{\sqrt{1 + 	an^2 	heta}}$,we have $\cos(	an^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$.Next,simplify the left-hand side: Let $\cot^{-1}(1 + x) = \phi$,then $1 + x = \cot \phi$. Since $\sin \phi = \frac{1}{\csc \phi} = \frac{1}{\sqrt{1 + \cot^2 \phi}}$,we have $\sin(\cot^{-1}(1 + x)) = \frac{1}{\sqrt{1 + (1 + x)^2}}$.Equating both sides: $\frac{1}{\sqrt{1 + (1 + x)^2}} = \frac{1}{\sqrt{1 + x^2}}$.Squaring both sides: $1 + (1 + x)^2 = 1 + x^2$.$(1 + x)^2 = x^2$.$1 + 2x + x^2 = x^2$.$1 + 2x = 0$.$x = -\frac{1}{2}$.

The value of $x$ for which $\sin(\cot^{-1} (1 + x)) = \cos(\tan^{-1} x)$ is

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