Question

The value of cot (sin⁻¹x) is
$\left(\mathrm{a}\right)\frac{\sqrt{1+x^2}}{x}\left(\mathrm{b}\right)\frac{x}{\sqrt{1+x^2}}\left(\mathrm{c}\right)\frac{1}{x}\left(\mathrm{d}\right)\frac{\sqrt{1-x^2}}{x}$

Answer 1

We know

$$\begin{gathered} {\sin }^{-1}x={\cot }^{-1}\frac{\sqrt{1-x^2}}{x} \\ \therefore \cot \left({\sin }^{-1}x\right)=\cot \left({\cot }^{-1}\frac{\sqrt{1-x^2}}{x}\right) \\ \Rightarrow \cot \left({\sin }^{-1}x\right)=\frac{\sqrt{1-x^2}}{x}\left[\cot \left({\cot }^{-1}x\right)=x\mathit{,}\forall x\in \mathrm{R}\right] \end{gathered}$$

Thus, the value of cot(sin⁻¹x) is $\frac{\sqrt{1-x^2}}{x}$.

Hence, the correct answer is option (d).