Question

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

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $-\sqrt{3}$
• D. $-\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

At $\theta=\frac{\pi}{6}, x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$
$\Rightarrow x=\log \left[\cot \left(\frac{\pi}{4}+\frac{\pi}{6}\right)\right]$
$\Rightarrow e^{x}=\frac{\cot \pi / 4 \cot \pi / 6-1}{\cot \pi / 6+\cot \pi / 4}$
$\Rightarrow e^{x}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\therefore e^{-x}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
$\because \sin \,h(x)=\frac{e^{x}-e^{-x}}{2}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}+1}-\frac{\sqrt{3}+1}{\sqrt{3}-1}}{2}$
$=\frac{(3+1-2 \sqrt{3})-(3+1+2 \sqrt{3})}{4}=-\frac{4 \sqrt{3}}{4}=-\sqrt{3}$