If theta = frac{pi}{12} and x = log left(cot | vedclass

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(A) Given: $x = \log \left(\cot \left(\frac{\pi}{4} + 	hetaight)ight)$Substitute $	heta = \frac{\pi}{12}$:$x = \log \left(\cot \left(\frac{\pi}{

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$\frac{2}{\sqrt{3}}$

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(A) Given: $x = \log \left(\cot \left(\frac{\pi}{4} + 	hetaight)ight)$Substitute $	heta = \frac{\pi}{12}$:$x = \log \left(\cot \left(\frac{\pi}{4} + \frac{\pi}{12}ight)ight) = \log \left(\cot \left(\frac{3\pi + \pi}{12}ight)ight) = \log \left(\cot \left(\frac{\pi}{3}ight)ight)$Since $\cot \left(\frac{\pi}{3}ight) = \frac{1}{\sqrt{3}}$,we have $x = \log \left(\frac{1}{\sqrt{3}}ight) = -\log \sqrt{3} = -\frac{1}{2} \log 3$.Then $e^x = \frac{1}{\sqrt{3}}$ and $e^{-x} = \sqrt{3}$.The definition of $\cosh x$ is $\frac{e^x + e^{-x}}{2}$.$\cosh x = \frac{\frac{1}{\sqrt{3}} + \sqrt{3}}{2} = \frac{\frac{1 + 3}{\sqrt{3}}}{2} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}}$.

If $\theta = \frac{\pi}{12}$ and $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$,then $\cosh x =$

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