Q78 / 170 · Manipal (MET) 2021 · Mathematics
$\cot ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$ is equal to
- Correct answerA$\frac{1-\sin \theta}{1+\sin \theta}$
- B$\frac{1-\cos \theta}{1+\cos \theta}$
- C$\frac{1+\sin \theta}{1-\sin \theta}$
- D$\frac{2-\sin \theta}{2+\sin \theta}$
Solution
$\cot ^2\left(\frac{\pi}{4}+\frac{\theta}{2}\right)$
$\begin{aligned} & =\left[\cot \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right]^2=\left[\frac{\cot \frac{\theta}{2} \cot \frac{\pi}{4}-1}{\cot \frac{\theta}{2}+\cot \frac{\pi}{4}}\right]^2 \\ & =\left[\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}\right]^2=\frac{1-\sin \theta}{1+\sin \theta}\end{aligned}$
Hence, the correct answer is option: 'A'.