Question

If ${\sec }^{-1}(\frac{1+x}{1-y})=a$, then $\frac{dy}{dx}$ is

• A. $\frac{y+1}{x-1}$
• B. $\frac{y-1}{x+1}$
• C. $\frac{x-1}{y+1}$
• D. $\frac{x-1}{y-1}$

Answer 1

Answer: (B)

Given , ${\sec }^{-1}\left(\frac{1+x}{1-y}\right)=a$
$\Rightarrow \frac{1+x}{1-y}=\sec a$
$\Rightarrow 1+x=\left(1-y\right)\sec a$
$\Rightarrow yseca=\sec a-1-x$
$\Rightarrow \frac{dy}{dx}\sec a=-1$
$\Rightarrow \frac{dy}{dx}=\frac{-1}{\sec a}=\frac{-1}{\left(\frac{1+x}{1-y}\right)}$ (from eq 1)
$\Rightarrow \frac{dy}{dx}=\frac{-\left(1-y\right)}{\left(1+x\right)}$
$\Rightarrow \frac{dy}{dx}=\frac{y-1}{x+1}$