JEE Main 2026 Jan 28 Shift 2, Mathematics Q6: Inverse Trigonometric Functions

Let $\theta ={\sin }^{-1}\frac{2}{\sqrt{13}}$ and $\varphi ={\cos }^{-1}\frac{3}{\sqrt{10}}$.

Then $\sin \theta =\frac{2}{\sqrt{13}},\cos \theta =\frac{3}{\sqrt{13}}\Rightarrow \tan \theta =\frac{2}{3}$.

And $\cos \varphi =\frac{3}{\sqrt{10}},\sin \varphi =\frac{1}{\sqrt{10}}\Rightarrow \tan \varphi =\frac{1}{3}$.

Using $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$: $\tan 2\theta =\frac{4/3}{1-4/9}=\frac{12}{5}$, and $\tan 2\varphi =\frac{2/3}{1-1/9}=\frac{3}{4}$.

$\tan (2\theta -2\varphi )=\frac{\tan 2\theta -\tan 2\varphi }{1+\tan 2\theta \tan 2\varphi }=\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot \frac{3}{4}}=\frac{33/20}{56/20}=\frac{33}{56}$.