JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Trigonometric Ratios and Identities

The four products regroup into an angle-difference and an angle-sum form. Collecting the sine terms, $\sin \frac{15\theta }{2}\cos 8\theta -\cos \frac{15\theta }{2}\sin 8\theta =\sin \left(\frac{15\theta }{2}-8\theta \right)=\sin \left(-\frac{\theta }{2}\right)$, and collecting the cosine terms, $\cos \frac{15\theta }{2}\cos 8\theta +\sin \frac{15\theta }{2}\sin 8\theta =\cos \left(\frac{15\theta }{2}-8\theta \right)=\cos \left(-\frac{\theta }{2}\right)$. Hence the expression equals $\cos \frac{\theta }{2}-\sin \frac{\theta }{2}$. Squaring gives ${\left(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right)}^2=1-\sin \theta$. Because $\frac{\pi }{2}<\theta <\pi$ forces $\frac{\pi }{4}<\frac{\theta }{2}<\frac{\pi }{2}$, where $\sin \frac{\theta }{2}>\cos \frac{\theta }{2}$, the quantity is negative, so $\cos \frac{\theta }{2}-\sin \frac{\theta }{2}=-\sqrt{1-\sin \theta }$. From $\cot \theta =-\frac{1}{2\sqrt{2}}$ with $\theta$ in the second quadrant, ${\csc }^2\theta =1+{\cot }^2\theta =\frac{9}{8}$, so $\sin \theta =\frac{2\sqrt{2}}{3}$. Then $1-\sin \theta =\frac{3-2\sqrt{2}}{3}$ and $\sqrt{3-2\sqrt{2}}=\sqrt{2}-1$, giving the value $-\frac{\sqrt{2}-1}{\sqrt{3}}=\frac{1-\sqrt{2}}{\sqrt{3}}$.