Mathematics › Trigonometric Functions › Sum and Difference & Multiple angle Formula
If $x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$ and $\theta \in\left(\frac{-\pi}{4},…
If $x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$ and $\theta \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$, then consider the following statements
I. $\cosh x=\sec 2 \theta$
II. $\sinh x=-\tan 2 \theta$
I is true and II is false I is false and II is true Both I and II are true Both I and II are false Solution
$
\begin{aligned}
& \text { We have, } x=\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right] \\
& \cosh x=\frac{e^x+e^{-x}}{2} \\
& =\frac{e^{\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right]}+e^{-\log \left[\cot \left(\frac{\pi}{4}+\theta\right)\right]}}{2} \\
& =\frac{\left\lceil\cot \left(\frac{\pi}{4}+\theta\right)+\frac{1}{\cot \left(\frac{\pi}{4}+\theta\right)}\right\rfloor}{2} \\
& =\frac{\cot ^2\left(\frac{\pi}{4}+\theta\right)+1}{2 \cot \left(\frac{\pi}{4}+\theta\right)}=\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}+\theta\right)}{2 \cot \left(\frac{\pi}{4}+\theta\right)} \\
& =\frac{1}{2 \sin ^2\left(\frac{\pi}{4}+\theta\right)} \cdot \frac{\sin \left(\frac{\pi}{4}+\theta\right)}{\cos \left(\frac{\pi}{4}+\theta\right)} \\
& =\frac{1}{2 \sin \left(\frac{\pi}{4}+\theta\right) \cos \left(\frac{\pi}{4}+\theta\right)} \\
& =\frac{1}{\sin \left(2\left(\frac{\pi}{4}+\theta\right)\right)}=\frac{1}{\sin \left(\frac{\pi}{2}+2 \theta\right)}=\frac{1}{\cos 2 \theta}=\sec 2 \theta \\
&
\end{aligned}
$
Hence, statement I is true.
Again, $\sinh x=\frac{e^x-e^{-x}}{2}$
$
\begin{aligned}
& =\frac{e^{\log \left(\cot \left(\frac{\pi}{4}+\theta\right)\right)}-e^{-\log \left(\cot \left(\frac{\pi}{4}+\theta\right)\right)}}{2} \\
& =\frac{\cot \left(\frac{\pi}{4}+\theta\right)-\frac{1}{\cot \left(\frac{\pi}{4}+\theta\right)}}{2} \\
& =\frac{\cot ^2\left(\frac{\pi}{4}+\theta\right)-1}{2 \cot \left(\frac{\pi}{4}+\theta\right)}=\frac{1-\tan ^2\left(\frac{\pi}{4}+\theta\right)}{2 \tan \left(\frac{\pi}{4}+\theta\right)} \\
& =\frac{1}{\left(\frac{\tan \left(\frac{\pi}{4}+\theta\right)}{1-\tan ^2\left(\frac{\pi}{4}+\theta\right)}\right)} \frac{\tan \left(2\left(\frac{\pi}{4}+\theta\right)\right)}{1} \\
& =\frac{1}{\tan \left(\frac{\pi}{2}+2 \theta\right)} \frac{1}{-\cot 2 \theta}=-\tan 2 \theta
\end{aligned}
$
$\therefore \quad$ statement II is also true
Asked in: AP EAMCET 2018 (23 Apr Shift 1)
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