Question

The value of $\sin \left[{\cot }^{-1}\left\{\cos \right({\tan }^{-1}x\left)\right\}\right]$ is

• A. $\left(\sqrt{\frac{1+x^2}{2+x^2}}\right)$
• B. $\left(\sqrt{\frac{2+x^2}{1+x^2}}\right)$
• C. $\left(\sqrt{\frac{x^2-2}{x^2-1}}\right)$
• D. $\left(\sqrt{\frac{x^2-1}{x^2-2}}\right)$

Answer 1

Answer: (A)

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

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

let $ta{n}^{-1}x=\theta$

$tan\theta =x$

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

the function is reduced to $sin\left[co{t}^{-1}a\right]$

let $co{t}^{-1}a={\theta }_2$

$cot{\theta }_2=a=\frac{1}{\sqrt{1+x^2}}$

$tan{\theta }_2=\sqrt{1+x^2}$

$sin{\theta }_2=\sqrt{\frac{1+x^2}{2+x^2}}$

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