If sec ^{-1}left(frac{1+x}{1-y}right)=a,then | vedclass

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(A) Given,$\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$ Taking $\sec$ on both sides,we get: $\frac{1+x}{1-y}=\sec a$ Rearranging the terms: $1+x = (1-y)

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$\frac{y-1}{x+1}$

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(A) Given,$\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$ Taking $\sec$ on both sides,we get: $\frac{1+x}{1-y}=\sec a$ Rearranging the terms: $1+x = (1-y) \sec a$ $1+x = \sec a - y \sec a$ $y \sec a = \sec a - 1 - x$ $y = \frac{\sec a - 1 - x}{\sec a} = 1 - \frac{1+x}{\sec a}$ Differentiating both sides with respect to $x$: $\frac{d y}{d x} = 0 - \frac{1}{\sec a} \cdot \frac{d}{d x}(1+x)$ $\frac{d y}{d x} = -\frac{1}{\sec a} \cdot (1)$ Since $\sec a = \frac{1+x}{1-y}$,we substitute this back: $\frac{d y}{d x} = -\frac{1}{\frac{1+x}{1-y}} = -\frac{1-y}{1+x} = \frac{y-1}{x+1}$

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

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