If $\cot x=\frac{5}{12}$ for some $x \in\left(\pi, \frac{3 \pi}{2}\right)$, then $\sin 7 x\left(\cos \frac{13 x}{2}+\sin \frac{13 x}{2}\right)+$ $\cos 7 x\left(\cos \frac{13 x}{2}-\sin \frac{13 x}{2}\right)$ is equal to

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Trigonometric Ratios and Identities

cot x = \frac{5}{12}

for some

x \in (π, \frac{3 π}{2})

, then

sin 7 x (cos \frac{13 x}{2} + sin \frac{13 x}{2}) +

cos 7 x (cos \frac{13 x}{2} - sin \frac{13 x}{2})

is equal to

Answer: (C)

Grouping the terms, the expression equals

[sin 7 x cos \frac{13 x}{2} - cos 7 x sin \frac{13 x}{2}] + [sin 7 x sin \frac{13 x}{2} + cos 7 x cos \frac{13 x}{2}] = sin (7 x - \frac{13 x}{2}) + cos (7 x - \frac{13 x}{2}) = sin \frac{x}{2} + cos \frac{x}{2}

. With

cot x = \frac{5}{12}

and

x \in (π, \frac{3 π}{2})

cos x = - \frac{5}{13}

. Since

\frac{x}{2} \in (\frac{π}{2}, \frac{3 π}{4})

sin \frac{x}{2} > 0

and

cos \frac{x}{2} < 0

: from

cos x = 1 - 2 sin^{2} \frac{x}{2}

sin \frac{x}{2} = \frac{3}{13}

, and from

cos x = 2 cos^{2} \frac{x}{2} - 1

cos \frac{x}{2} = - \frac{2}{13}

. Hence the value is

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

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Single correct option