If $ heta=\frac{\pi}{6}$ and $x=\log \left[\cot \left(\frac{\pi}{4}+ heta ight) ight]$, then $\sin h(x)=$

Question

If $	heta=\frac{\pi}{6}$ and $x=\log \left[\cot \left(\frac{\pi}{4}+	hetaight)ight]$, then $\sin h(x)=$

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { At } 	heta=\frac{\pi}{6}, x=\log \left[\cot \left(\frac{\pi}{4}+	hetaight)ight] \ & \Rightarrow \quad x=\log \left[\cot \left(\frac{\pi}{4}+\frac{\pi}{6}ight)ight] \ & \Rightarrow \quad e^x=\frac{\cot \pi / 4 \cot \pi / 6-1}{\cot \pi / 6+\cot \pi / 4} \ & \Rightarrow \quad e^x=\frac{\sqrt{3}-1}{\sqrt{3}+1} \ & 	herefore \quad e^{-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \ & \because \quad \sin h(x)=\frac{e^x-e^{-x}}{2}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}+1}-\frac{\sqrt{3}+1}{\sqrt{3}-1}}{2} \ & =\frac{(3+1-2 \sqrt{3})-(3+1+2 \sqrt{3})}{4}=-\frac{4 \sqrt{3}}{4}=-\sqrt{3} \ & \end{aligned} $ Hence, option (c) is correct

If $\theta=\frac{\pi}{6}$ and $x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$, then $\sin h(x)=$

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