Q64 / 160 · AP EAMCET 2023 (18 May Shift 2) · Mathematics
If $\cosh x=\operatorname{cosec} \theta$, then $\operatorname{coth} 2 \frac{x}{2}=$
- A$\tan ^2 \frac{\theta}{2}$
- B$\tan ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
- C$\cot ^2 \frac{\theta}{2}$
- Correct answerD$\cot ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
Solution
$\begin{aligned} & \text {Since } \cos h x=\operatorname{cosec} \theta \\ & \Rightarrow \frac{1+\tan -h^2 \frac{x}{2}}{1-\tan ^2 \frac{x}{2}}=\operatorname{cosec} \theta \Rightarrow \frac{2}{2 \tan h^2 \frac{x}{2}}=\frac{\operatorname{cosec} \theta+1}{\operatorname{cosec} \theta-1} \\ & \Rightarrow \cot h^2 \frac{x}{2}=\frac{1+\sin \theta}{1-\sin \theta} \\ & \Rightarrow \operatorname{coth}^2 \frac{x}{2}=\frac{\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)^2}{\left(\sin ^2 \frac{\theta}{2}-\cos \frac{\theta}{2}\right)^2}=\left(\frac{\cot \frac{\theta}{2}+1}{\cot \frac{\theta}{2}-1}\right)^2\end{aligned}$
$=\left(\frac{\cot \frac{\theta}{2} \cot \frac{\pi}{4}+1}{\cot \frac{\theta}{2}-\cot \frac{\theta}{4}}\right)^2=\cot ^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)$
Hence, the correct answer is option: 'D'.