If x=log _eleft(frac{cos frac{y}{2}-sin frac{ | vedclass

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(B) Given $x=\log _{e}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$.Dividing numerator and denominator by

Vedclass · Accepted Answer

$-\frac{1}{2}$

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(B) Given $x=\log _{e}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}ight)$.Dividing numerator and denominator by $\cos \frac{y}{2}$,we get $e^x=\frac{1-	an \frac{y}{2}}{1+	an \frac{y}{2}} = 	an \left(\frac{\pi}{4}-\frac{y}{2}ight) \quad \dots(i)$.Differentiating both sides with respect to $x$,we get $e^x = \sec^2 \left(\frac{\pi}{4}-\frac{y}{2}ight) \cdot \left(-\frac{1}{2}ight) \frac{dy}{dx}$.Thus,$\frac{dy}{dx} = -2e^x \cos^2 \left(\frac{\pi}{4}-\frac{y}{2}ight)$.When $t=\frac{1}{2}$,$	an \frac{y}{2} = \sqrt{\frac{1-1/2}{1+1/2}} = \sqrt{\frac{1/2}{3/2}} = \frac{1}{\sqrt{3}}$.So,$\frac{y}{2} = \frac{\pi}{6}$.Substituting $\frac{y}{2} = \frac{\pi}{6}$ into equation $(i)$,$e^x = 	an \left(\frac{\pi}{4}-\frac{\pi}{6}ight) = 	an \frac{\pi}{12} = 2-\sqrt{3}$.Now,$\cos^2 \left(\frac{\pi}{4}-\frac{\pi}{6}ight) = \cos^2 \frac{\pi}{12} = \left(\frac{\sqrt{3}+1}{2\sqrt{2}}ight)^2 = \frac{3+1+2\sqrt{3}}{8} = \frac{4+2\sqrt{3}}{8} = \frac{2+\sqrt{3}}{4}$.Therefore,$\left(\frac{dy}{dx}ight)_{t=\frac{1}{2}} = -2(2-\sqrt{3}) \cdot \frac{2+\sqrt{3}}{4} = -\frac{1}{2}(4-3) = -\frac{1}{2}$.

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

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