Q101 / 180 · COMEDK 2020 · Mathematics

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

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

Solution

We have, $$ \begin{gathered} \sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \frac{1+x}{1-y}=\sec a \\ \Rightarrow \quad \frac{1(1-y)-(1+x)\left(-\frac{d y}{d x}\right)}{(1-y)^{2}}=0 \\ \Rightarrow \quad 1-y+(1+x) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1} \end{gathered} $$

Hence, the correct answer is option: 'B'.

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