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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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1027 Upvotes
Verified Answer
The correct answer is:
$\frac{y-1}{x+1}$
We have,
$$
\begin{gathered}
\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \frac{1+x}{1-y}=\sec a \\
\Rightarrow \quad \frac{1(1-y)-(1+x)\left(-\frac{d y}{d x}\right)}{(1-y)^{2}}=0 \\
\Rightarrow \quad 1-y+(1+x) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}
\end{gathered}
$$
$$
\begin{gathered}
\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \frac{1+x}{1-y}=\sec a \\
\Rightarrow \quad \frac{1(1-y)-(1+x)\left(-\frac{d y}{d x}\right)}{(1-y)^{2}}=0 \\
\Rightarrow \quad 1-y+(1+x) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}
\end{gathered}
$$
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