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

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(D) Given the equation: $\sin \left(\cot ^{-1}(x+1)ight)=\cos \left(	an ^{-1} xight)$ Let $	heta_1 = \cot ^{-1}(x+1)$. Then $\cot 	heta_1 = x+1

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$-\frac{1}{2}$

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(D) Given the equation: $\sin \left(\cot ^{-1}(x+1)ight)=\cos \left(	an ^{-1} xight)$ Let $	heta_1 = \cot ^{-1}(x+1)$. Then $\cot 	heta_1 = x+1$. Using the identity $\sin 	heta = \frac{1}{\sqrt{1+\cot^2 	heta}}$,we get $\sin 	heta_1 = \frac{1}{\sqrt{1+(x+1)^2}} = \frac{1}{\sqrt{1+x^2+2x+1}} = \frac{1}{\sqrt{x^2+2x+2}}$. Let $	heta_2 = 	an ^{-1} x$. Then $	an 	heta_2 = x$. Using the identity $\cos 	heta = \frac{1}{\sqrt{1+	an^2 	heta}}$,we get $\cos 	heta_2 = \frac{1}{\sqrt{1+x^2}}$. Equating the two expressions: $\frac{1}{\sqrt{x^2+2x+2}} = \frac{1}{\sqrt{1+x^2}}$. 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 $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$,then considering positive square roots,$x$ has the value $\qquad$

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