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 y sec a = \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} $