Question

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

• A. $\frac{x}{\sqrt{x^2+2}}$
• B. $\frac{x}{\sqrt{x^2+1}}$
• C. $\frac{1}{\sqrt{x^2+2}}$
• D. $\sqrt{\frac{x^2+1}{x^2+2}}$

Answer 1

The correct option is D

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

Explanation for the correct answer:

Simplifying the equations using trigonometric identities:

Let ${\tan }^{-1}\left(x\right)=y$

$\therefore x=\tan y$

thus, $\cos \left({\tan }^{-1}\left(x\right)\right)=\cos y$

$$\begin{gathered} =\frac{1}{\sec \left(y\right)} \\ =\frac{1}{\sqrt{1+{\tan }^{2}y}}\left[\because se{c}^{2}x=1+{\tan }^{2}x\right] \\ =\frac{1}{\sqrt{1+x^2}} \end{gathered}$$

Let, $t={\cot }^{-1}\left(\cos \left({\tan }^{-1}\left(x\right)\right)\right)$

Putting the value of $\cos \left({\tan }^{-1}\left(x\right)\right)=\frac{1}{\sqrt{1+x^2}}$

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

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

We know that, ${\csc }^{2}t=1+co{t}^{2}t$, thus putting the value of $cott$

$$\begin{gathered} \therefore {\csc }^{2}t=1+{\left(\frac{1}{\sqrt{1+x^2}}\right)}^2 \\ \Rightarrow {\csc }^{2}t=1+\left(\frac{1}{1+x^2}\right) \\ \Rightarrow {\csc }^{2}t=\frac{1+x^2+1}{1+x^2} \\ \Rightarrow {\csc }^{2}t=\frac{x^2+2}{x^2+1} \\ \Rightarrow {\sin }^{2}t=\frac{x^2+1}{x^2+2}\mathbf{}\left[\because \sin \left(x\right)=\frac{1}{\csc \left(x\right)}\right] \\ \Rightarrow \sin t=\sqrt{\frac{x^2+1}{x^2+2}} \end{gathered}$$

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

Therefore, the correct answer is option (D).