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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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Verified Answer
The correct answer is:
$-\sqrt{3}$
$$
\begin{aligned}
& \text { At } \theta=\frac{\pi}{6}, x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\
& \Rightarrow \quad x=\log \left[\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)\right] \\
& \Rightarrow \quad e^x=\frac{\cot \pi / 4 \cot \pi / 6-1}{\cot \pi / 6+\cot \pi / 4} \\
& \Rightarrow \quad e^x=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\
& \therefore \quad e^{-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \\
& \because \quad \sin h(x)=\frac{e^x-e^{-x}}{2}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}+1}-\frac{\sqrt{3}+1}{\sqrt{3}-1}}{2} \\
& =\frac{(3+1-2 \sqrt{3})-(3+1+2 \sqrt{3})}{4}=-\frac{4 \sqrt{3}}{4}=-\sqrt{3} \\
&
\end{aligned}
$$
Hence, option (c) is correct.
\begin{aligned}
& \text { At } \theta=\frac{\pi}{6}, x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\
& \Rightarrow \quad x=\log \left[\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)\right] \\
& \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} \\
& \therefore \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.
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