JEE Main 2026 — 24 January, Morning Shift — Question 10

$\cot x=\frac{5}{12} \Rightarrow \cos x=\frac{-5}{13}=2 \cos ^{2} \frac{x}{2}-1$

$\cos \left(\frac{x}{2}\right)=-\frac{2}{\sqrt{13}}$ or $\frac{2}{\sqrt{13}}($ rejected $)$

$\left\{\because \frac{\mathrm{x}}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)\right\}$

$\left(\sin 7 \mathrm{x} \frac{\sin 13 \mathrm{x}}{2}+\cos 7 \mathrm{x} \frac{\cos 13 \mathrm{x}}{2}\right)+\left(\sin 7 \mathrm{x} \frac{\cos 13 \mathrm{x}}{2}-\cos 7 \mathrm{x} \frac{\sin 13 \mathrm{x}}{2}\right)$

$\cos \left(7 \mathrm{x}-\frac{13 \mathrm{x}}{2}\right)+\sin \left(7 \mathrm{x}-\frac{13 \mathrm{x}}{2}\right)$

$\cos \frac{\mathrm{x}}{2}+\sin \left(\frac{\mathrm{x}}{2}\right)$

$\frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}=\frac{1}{\sqrt{13}}$