If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}} ight), an \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$. Then $\left(y_1 ight)_{t=\frac{1}{2}}$ has the value

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If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}ight), 	an \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$. Then $\left(y_1ight)_{t=\frac{1}{2}}$ has the value

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$\begin{aligned} & x=\log _{\mathrm{e}}\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}ight) \ & \Rightarrow \mathrm{e}^x=\frac{1-	an \frac{y}{2}}{1+	an \frac{y}{2}}\end{aligned}$ $\Rightarrow \mathrm{e}^x=	an \left(\frac{\pi}{4}-\frac{y}{2}ight) \quad$....(i) $\left[\because 	an \frac{\pi}{4}=1ight]$ Differentiating w.r.t. $x$, we get $\mathrm{e}^x=\sec ^2\left(\frac{\pi}{4}-\frac{y}{2}ight) \cdot\left(\frac{-1}{2}ight) \frac{\mathrm{d} y}{\mathrm{~d} x}$ $\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-2 \mathrm{e}^x \cos ^2\left(\frac{\pi}{4}-\frac{y}{2}ight)$ $\begin{aligned} & 	ext { When } \mathrm{t}=\frac{1}{2}, \ & 	an \frac{y}{2}=\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}}} \ & \Rightarrow 	an \frac{y}{2}=\frac{1}{\sqrt{3}} \ & \Rightarrow \frac{y}{2}=\frac{\pi}{6}\end{aligned}$ Substituting $\frac{y}{2}=\frac{\pi}{6}$ in (i), we get $\mathrm{e}^x=	an \frac{\pi}{12}=2-\sqrt{3}$ $	herefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{\mathrm{t}=\frac{1}{2}}=-2(2-\sqrt{3}) \cos ^2 \frac{\pi}{12}$ $\begin{aligned} & =-2(2-\sqrt{3})\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}ight)^2 \ & =\frac{-1}{2}(2-\sqrt{3})(2+\sqrt{3}) \ & =-\frac{1}{2}\end{aligned}$

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right), \tan…

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