The value of x which satisfies sin left(cot ^ | vedclass

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Question

(A) Given equation is $\sin \left(\cot ^{-1} xight)=\cos \left(	an ^{-1}(1+x)ight)$.We know that $\sin \left(\cot ^{-1} xight) = \sin \left(\si

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$-\frac{1}{2}$

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(A) Given equation is $\sin \left(\cot ^{-1} xight)=\cos \left(	an ^{-1}(1+x)ight)$.We know that $\sin \left(\cot ^{-1} xight) = \sin \left(\sin ^{-1} \frac{1}{\sqrt{1+x^2}}ight) = \frac{1}{\sqrt{1+x^2}}$.Also,$\cos \left(	an ^{-1}(1+x)ight) = \cos \left(\cos ^{-1} \frac{1}{\sqrt{1+(1+x)^2}}ight) = \frac{1}{\sqrt{1+(1+x)^2}}$.Equating both sides,we get $\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+(1+x)^2}}$.Squaring both sides,$1+x^2 = 1+(1+x)^2$.$1+x^2 = 1+1+x^2+2x$.$1+x^2 = 2+x^2+2x$.Subtracting $x^2$ from both sides,$1 = 2+2x$.$2x = -1$.$x = -\frac{1}{2}$.

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

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