Question

If $x={\cos }^{-1}\frac{1}{\sqrt{1+t^2}}\mathrm{and}y={\sin }^{-1}\frac{t}{\sqrt{1+t^2}}$, then $\frac{dy}{dx}$ is equal to:

• A. $0$
• B. $\tan t$
• C. $1$
• D. $\sin t\cos t$

Answer 1

The correct option is C

$1$

Explanation for the correct option.

$x={\cos }^{-1}\frac{1}{\sqrt{1+t^2}}$

Let $t=\tan \theta$, then

$\begin{array}{rcl}x& =& {\cos }^{-1}\frac{1}{\sqrt{1+{\tan }^2\theta }}\\ & =& {\cos }^{-1}\frac{1}{\sec \theta }\\ & =& {\cos }^{-1}\left(\cos \theta \right)\\ & =& \theta \end{array}$

$\begin{array}{rcl}& \Rightarrow & \frac{dx}{d\theta }=1\end{array}$

Similarly,

$\begin{array}{rcl}y& =& {\sin }^{-1}\frac{\tan \theta }{\sqrt{1+{\tan }^2\theta }}\\ & =& {\sin }^{-1}\frac{\tan \theta }{\sec \theta }\\ & =& {\sin }^{-1}\left(\sin \theta \right)\\ & =& \theta \end{array}$

$\begin{array}{rcl}& \Rightarrow & \frac{dy}{d\theta }=1\end{array}$

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & \frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}\end{array} \\ =1 \end{gathered}$$

Hence, option C is correct.