Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$

Question

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?

Answer 1

$\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x)).$ 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 $LHS=RHS \implies 2+2x+x^2=1+x^2 \implies x=-1/2.$

Answer 2

$\sin(\cot^{-1}(1+x))=\cos\left(\dfrac\pi2-\cot^{-1}(1+x)\right)=\cos(\tan^{-1}(1+x))$

We need $$\tan^{-1}(1+x)=2n\pi\pm\tan^{-1}x$$ where $n$ is any integer

Considering the '-' sign, $$\tan^{-1}(1+x)+\tan^{-1}x=2n\pi$$

Using Principal values, $$-\pi<\tan^{-1}(1+x)+\tan^{-1}x<\pi\implies n=0$$

Consequently, $$\tan^{-1}(1+x)=-\tan^{-1}x=\tan^{-1}(-x)$$

$$\iff1+x=-x\iff x=?$$

Check for the '+' sign