JEE Main 2026 (Online) 24th January Morning ShiftQuestion 1
If $\cot x=\frac{5}{12}$ for some $x\in\left(\pi,\frac{3\pi}{2}\right)$, then $\sin7x\left(\cos\frac{13x}{2}+\sin\frac{13x}{2}\right)+\cos7x\left(\cos\frac{13x}{2}-\sin\frac{13x}{2}\right)$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Regroup the four products into $\sin(7x-\frac{13x}2)$ and $\cos(7x-\frac{13x}2)$. Formulas used. (1) $\sin P\cos Q-\cos P\sin Q=\sin(P-Q)$, $\cos P\cos Q+\sin P\sin Q=\cos(P-Q)$(2) $\sin^2\frac x2=\frac{1-\cos x}{2}$, $\cos^2\frac x2=\frac{1+\cos x}{2}$Step 1 - Regroup. With $P=7x$, $Q=\frac{13x}{2}$, $P-Q=\frac x2$: $$\text{expression}=(\sin P\cos Q-\cos P\sin Q)+(\cos P\cos Q+\sin P\sin Q)=\sin\frac x2+\cos\frac x2.$$ Step 2 - $\cos x$. $\cot x=\frac5{12}$ in the third quadrant gives $\cos x=-\frac5{13}$. Step 3 - Half angle. $\frac x2\in\left(\frac\pi2,\frac{3\pi}4\right)$, so $\sin\frac x2>0$ and $\cos\frac x2<0$. By (2): $$\sin\frac x2=\sqrt{\frac{18}{26}}=\frac{3}{\sqrt{13}},\qquad\cos\frac x2=-\sqrt{\frac{8}{26}}=-\frac{2}{\sqrt{13}}.$$ Step 4 - Value. $\dfrac{3-2}{\sqrt{13}}=\dfrac{1}{\sqrt{13}}$. Answer: (A)