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