For -frac{pi}{2}

Question

(B) Let $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$.Using the trigonometric identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) Let $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$.Using the trigonometric identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$:$\frac{\cos x}{1 + \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2})}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = 	an(\frac{\pi}{4} - \frac{x}{2})$.Thus, $y = 	an^{-1} \left( 	an(\frac{\pi}{4} - \frac{x}{2}) ight) = \frac{\pi}{4} - \frac{x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} ight) = -\frac{1}{2}$.

For $-\frac{\pi}{2} < x < \frac{3 \pi}{2}$, the value of $\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}$ is equal to

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