JEE Main 2026 — 28 January, Evening Shift — Question 6

Let $\sin ^{-1} \frac{2}{\sqrt{13}}=\theta , \cos ^{-1} \frac{3}{\sqrt{10}}=\phi$

$\sin \theta=\frac{2}{\sqrt{13}} \& \cos \phi=\frac{3}{\sqrt{10}}$

$\tan (2 \theta-2 \phi)=\frac{\tan 2 \theta-\tan 2 \phi}{1+\tan 2 \theta \tan 2 \phi}$

$\left(\because \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)$

$=\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5} \cdot \frac{3}{4}}$

$=\frac{33}{56}$