MCQ
If $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$, then $\frac{d y}{d x}$ is equal to
  • A
    $\frac{x-1}{y-1}$
  • B
    $\frac{x-1}{y+1}$
  • C
    $\frac{y-1}{x+1}$
  • D
    $\frac{y+1}{x-1}$

Answer

Given, $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \sec a=\frac{1+x}{1-y}$
On differentiating, we get
$\frac{(1-y)+(1+x) \frac{d y}{d x}}{(1-y)^2}=0 \Rightarrow(1+x) \frac{d y}{d x}=y-1 \Rightarrow \frac{d y}{d x}=\frac{y-1}{1+x}
$

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