Question

$If\sin \left[{\cot }^{-1}\left(x+1\right)\right]=\cos \left({\tan }^{-1}x\right),thenfindx$

• A. $\frac{-1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. None of these

Answer 1

The correct option is A $\frac{-1}{2}$

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

${\cot }^{-1}\left(x+1\right)=p$

$\cot p=x+1$

$\sin p=\frac{1}{\sqrt{1+{\left(x+1\right)}^2}}=\frac{1}{\sqrt{x^2+2x+2}}$

$p={\sin }^{-1}\left(\frac{1}{\sqrt{x^2+2x+2}}\right)$

& ${\tan }^{-1}x=q$

$x=\tan q$

$\cos q=\frac{1}{\sqrt{1+x^2}}$

$q={\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

(1) Becomes

$\sin \left[{\sin }^{-1}\left(\frac{1}{\sqrt{x^2+2x+2}}\right)=\cos \left[{\cos }^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right]\right]$

$\Rightarrow \frac{1}{\sqrt{x^2+2x+2}}=\frac{1}{\sqrt{x^2+1}}$$\Rightarrow x^2+1=x^2+2x+2$

$\Rightarrow 2x=-1$

$x=-\frac{1}{2}$