Q31 / 75 · JEE Main 2026 (24 January Shift 1) · Mathematics
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
- A$\frac{4}{\sqrt{26}}$
- B$\frac{6}{\sqrt{26}}$
- C$\frac{5}{\sqrt{13}}$
- Correct answerD$\frac{1}{\sqrt{13}}$
Solution
$\cot x = \dfrac{5}{12} \Rightarrow \cos x = \dfrac{-5}{13} = 2\cos^2 \dfrac{x}{2} - 1$
$\cos\left(\dfrac{x}{2}\right) = -\dfrac{2}{\sqrt{13}}$ or $\dfrac{2}{\sqrt{13}}$ (rejected)
$\left\{ \because \dfrac{x}{2} \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{4}\right) \right\}$
$\left(\sin 7x \dfrac{\sin 13x}{2} + \cos 7x \dfrac{\cos 13x}{2}\right) + \left(\sin 7x \dfrac{\cos 13x}{2} - \cos 7x \dfrac{\sin 13x}{2}\right)$
$\cos\left(7x - \dfrac{13x}{2}\right) + \sin\left(7x - \dfrac{13x}{2}\right)$
$\cos \dfrac{x}{2} + \sin\left(\dfrac{x}{2}\right)$
$\dfrac{3}{\sqrt{13}} - \dfrac{2}{\sqrt{13}} = \dfrac{1}{\sqrt{13}}$
$\cos\left(\dfrac{x}{2}\right) = -\dfrac{2}{\sqrt{13}}$ or $\dfrac{2}{\sqrt{13}}$ (rejected)
$\left\{ \because \dfrac{x}{2} \in \left(\dfrac{\pi}{2}, \dfrac{3\pi}{4}\right) \right\}$
$\left(\sin 7x \dfrac{\sin 13x}{2} + \cos 7x \dfrac{\cos 13x}{2}\right) + \left(\sin 7x \dfrac{\cos 13x}{2} - \cos 7x \dfrac{\sin 13x}{2}\right)$
$\cos\left(7x - \dfrac{13x}{2}\right) + \sin\left(7x - \dfrac{13x}{2}\right)$
$\cos \dfrac{x}{2} + \sin\left(\dfrac{x}{2}\right)$
$\dfrac{3}{\sqrt{13}} - \dfrac{2}{\sqrt{13}} = \dfrac{1}{\sqrt{13}}$
Hence, the correct answer is option: 'D'.