JEE Main 2026 Jan 24 Shift 1, Mathematics Q10: Trigonometric Ratios And Identities

Simplify the expression first. Group by the half-angle sums:

$E=\cos \frac{13x}{2}(\sin 7x+\cos 7x)+\sin \frac{13x}{2}(\sin 7x-\cos 7x)$

$=\left[\sin 7x\cos \frac{13x}{2}-\cos 7x\sin \frac{13x}{2}\right]+\left[\cos 7x\cos \frac{13x}{2}+\sin 7x\sin \frac{13x}{2}\right]$

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

Find $\sin (x/2)$ and $\cos (x/2)$. In Q3 ($\pi <x<3\pi /2$), both $\sin x$ and $\cos x$ are negative. From $\cot x=5/12$: $\cos x=-5/13$, $\sin x=-12/13$.

Since $x/2\in (\pi /2,3\pi /4)$ (Q2): $\sin (x/2)>0$, $\cos (x/2)<0$.

$\cos x=2{\cos }^2(x/2)-1=-\frac{5}{13}\Rightarrow {\cos }^2(x/2)=\frac{4}{13}\Rightarrow \cos (x/2)=-\frac{2}{\sqrt{13}}$.

${\sin }^2(x/2)=1-\frac{4}{13}=\frac{9}{13}\Rightarrow \sin (x/2)=\frac{3}{\sqrt{13}}$.

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