Question

$cos[ta{n}^{-1}\{sin(co{t}^{-1}x)\}]$ 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

Answer: (C)

We have, $cos[ta{n}^{-1}\{sin(co{t}^{-1}x)\}]$


Let $co{t}^{-1}x=\theta \Rightarrow cot\theta =x$
$\Rightarrow sin\theta =\frac{1}{\sqrt{1+x^2}}$
$\therefore cos\left[ta{n}^{-1}\left\{sin\theta \right\}\right]=cos\left[ta{n}^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right]$
Again, let $ta{n}^{-1}\frac{1}{\sqrt{1+x^2}}=\varphi$
$\Rightarrow tan\varphi =\frac{1}{\sqrt{1+x^2}}$



$\Rightarrow cos\varphi =\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}$
$\therefore cos\left[ta{n}^{-1}\frac{1}{\sqrt{1+x^2}}\right]=cos\varphi$
$=\sqrt{\frac{x^2+1}{x^2+2}}$