Question

$cos\left[ta{n}^{-1}\left\{sin\left(co{t}^{-1}x\right)\right\}\right]$ is equal to

• A.
  
• B.
  
• C.
  
• D. x

Answer 1

The correct option is C

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