<s>[INST] Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$ Let $cot^{-1} (1+x)=a$
Then $x =1-\cot a$
So $$\sin a =\cos (\tan^{1}(1-\cot a))$$
$$a=\frac{\pi}{2}- \tan^{-1} (1-\cot a)$$
$$\cot a =\frac 12$$
So $x=\frac 12$, but given answer is $-\frac 12$. What am I doing wrong?
 [/INST] $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x)).$
Let $\cot^{-1} (1+x)=A \implies \cot A =1+x.$ So $LHS=\sin A= 
\frac{1}{\sqrt{2+2x+x^2}}$ Let $\tan B=x \implies RHS=\cos B =\frac{1}{\sqrt{1+x^2}}$. Finally $LHS=RHS \implies 2+2x+x^2=1+x^2 \implies x=-1/2.$
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