JEE Main 2026 — 23 January, Evening Shift — Question 2

$\frac{\pi}{2}<\theta<\pi$ and $\cot \theta=-\frac{1}{2 \sqrt{2}}$

$\Rightarrow \sin \left(\frac{15 \theta}{2}\right)(\cos 8 \theta+\sin 8 \theta)$

$+\cos \left(\frac{15 \theta}{2}\right)(\cos 8 \theta-\sin 8 \theta)$

$\Rightarrow \sin \left(\frac{15 \theta}{2}\right) \cos 8 \theta-\cos \left(\frac{15 \theta}{2}\right) \sin 8 \theta$ $+\sin \frac{15 \theta}{2} \sin 8 \theta+\cos \frac{15 \theta}{2} \cos 8 \theta$

$\Rightarrow \sin \left(\frac{15 \theta}{2}-8 \theta\right)+\cos \left(\frac{15 \theta}{2}-8 \theta\right)$

$\Rightarrow \cos \frac{\theta}{2}-\sin \frac{\theta}{2}=-\sqrt{1-\sin \theta} \quad\left(\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{2}\right)$

⇒ given $\cot \theta=-\frac{1}{2 \sqrt{2}}, \sin \theta=\frac{2 \sqrt{2}}{3}$

$\Rightarrow-\sqrt{1-\sin \theta}=\sqrt{\frac{3-2 \sqrt{2}}{3}}=-\frac{(\sqrt{2}-1)}{3}$ $=\frac{1-\sqrt{2}}{\sqrt{3}}$