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If $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$, then $\frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$\frac{y-1}{x+1}$
Given, $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$
$\Rightarrow \quad \frac{1+x}{1-y}=\sec a$
$\Rightarrow \quad 1+x=(1-y) \sec a$
$\Rightarrow \quad y \sec a=\sec a-1-x$
$\Rightarrow \quad \frac{d y}{d x} \sec a=-1$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-1}{\sec a}=\frac{-1}{\left(\frac{1+x}{1-y}\right)}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-(1-y)}{(1+x)} \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}$
$\Rightarrow \quad \frac{1+x}{1-y}=\sec a$
$\Rightarrow \quad 1+x=(1-y) \sec a$
$\Rightarrow \quad y \sec a=\sec a-1-x$
$\Rightarrow \quad \frac{d y}{d x} \sec a=-1$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-1}{\sec a}=\frac{-1}{\left(\frac{1+x}{1-y}\right)}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{-(1-y)}{(1+x)} \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}$
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