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

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(C) We know that $\sin \left(\cot ^{-1} x\right) = \sin \left(\sin ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right)\right) = \frac{1}{\sqrt{1+x^2}}$.Also,$\c

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$-\frac{1}{2}$

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

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

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