For $0 < x \leq \pi, \sinh ^{-1}(\cot x)$ is equal to
For $0 < x \leq \pi, \sinh ^{-1}(\cot x)$ is equal to
- $\log \left(\cot \frac{x}{2}\right)$
- $\log \left(\tan \frac{x}{2}\right)$
- $\log (1+\cot x)$
- $\log (1+\tan x)$
Solution
We know that
$\sinh ^{-1}(y)=\log \left(y+\sqrt{1+y^2}\right)$
Put
$y=\cot x$
$\begin{aligned}
\Rightarrow \sinh ^{-1}(\cot x) & =\log \left(\cot x+\sqrt{1+\cot ^2 x}\right) \\
& =\log \left(\cot x+\sqrt{\operatorname{cosec}^2 x}\right) \\
& =\log (\cot x+\operatorname{cosec} x) \\
& =\log \left(\frac{1+\cos x}{\sin x}\right)
\end{aligned}$
$\begin{aligned} & =\log \left(\frac{2 \cos ^2 x / 2}{2 \sin x / 2 \cdot \cos x / 2}\right) \\ & =\log (\cot x / 2)\end{aligned}$
Asked in: AP EAMCET 2011
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