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 side: Let $	an^{-1} x = 	heta$,then $x = 	an

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

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

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

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