If theta = frac{pi}{6} and x = log left[ cot | vedclass

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(C) Given $	heta = \frac{\pi}{6}$,we have $x = \log \left[ \cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) ight]$.$x = \log \left[ \cot \left( \f

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$-\sqrt{3}$

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(C) Given $	heta = \frac{\pi}{6}$,we have $x = \log \left[ \cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) ight]$.$x = \log \left[ \cot \left( \frac{5\pi}{12} ight) ight]$.Using the formula $\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$,we get $\cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) = \frac{\cot(\pi/4) \cot(\pi/6) - 1}{\cot(\pi/6) + \cot(\pi/4)} = \frac{1 \cdot \sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.Thus,$e^x = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ and $e^{-x} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.Using the definition $\sinh(x) = \frac{e^x - e^{-x}}{2}$,we have $\sinh(x) = \frac{1}{2} \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} - \frac{\sqrt{3}+1}{\sqrt{3}-1} ight)$.$\sinh(x) = \frac{1}{2} \left( \frac{(\sqrt{3}-1)^2 - (\sqrt{3}+1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} ight) = \frac{1}{2} \left( \frac{(3+1-2\sqrt{3}) - (3+1+2\sqrt{3})}{3-1} ight) = \frac{1}{2} \left( \frac{-4\sqrt{3}}{2} ight) = -\sqrt{3}$.Therefore,the correct option is $C$.

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

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