Question

$\cos \left[{\tan }^{-1}\right\{\sin \left({\cot }^{-1}x\right)\left\}\right]$ is equal to:

• A. $\sqrt{\frac{x^2+2}{x^2+3}}$
• B. $\sqrt{\frac{x^2+2}{x^2+1}}$
• C. $\sqrt{\frac{x^2+1}{x^2+2}}$
• D. none of these

Answer 1

The correct option is C $\sqrt{\frac{x^2+1}{x^2+2}}$

Let ${\cot }^{-1}x=\theta \Rightarrow \cot \theta =x$

Now, $\sin \theta =\frac{1}{\sqrt{{\cot }^2\theta +1}}$

So, $\sin \theta =\frac{1}{\sqrt{x^2+1}}\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)$

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

Again, Let ${\tan }^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)=\alpha \Rightarrow \tan \alpha =\left(\frac{1}{\sqrt{x^2+1}}\right)$

Now, $\cos \alpha =\frac{1}{\sqrt{{\tan }^2\alpha +1}}$

So, $\cos \alpha =\frac{1}{\sqrt{(\frac{1}{\sqrt{x^2+1}}{)}^2+1}}\Rightarrow \cos \alpha =\sqrt{\frac{x^2+1}{x^2+2}}\Rightarrow \alpha ={\cos }^{-1}\sqrt{\frac{x^2+1}{x^2+2}}$

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