Inverse Trigonometric Functions – Evaluation – JEE Main 28 Jan 2026 Shift 2

Solution:

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

We convert these to tangent:
For $\theta$: $\sin \theta = \frac{2}{\sqrt{13}}$.
$\text{Base} = \sqrt{(\sqrt{13})^2 – 2^2} = \sqrt{13-4} = \sqrt{9} = 3$.
$\Rightarrow \tan \theta = \frac{2}{3}$.

For $\phi$: $\cos \phi = \frac{3}{\sqrt{10}}$.
$\text{Perpendicular} = \sqrt{(\sqrt{10})^2 – 3^2} = \sqrt{10-9} = 1$.
$\Rightarrow \tan \phi = \frac{1}{3}$.

We need to find $\tan(2\theta – 2\phi)$.
First, calculate $\tan 2\theta$ and $\tan 2\phi$:
$$\tan 2\theta = \frac{2 \tan \theta}{1 – \tan^2 \theta} = \frac{2(2/3)}{1 – (2/3)^2} = \frac{4/3}{1 – 4/9} = \frac{4/3}{5/9} = \frac{4}{3} \times \frac{9}{5} = \frac{12}{5}$$
$$\tan 2\phi = \frac{2 \tan \phi}{1 – \tan^2 \phi} = \frac{2(1/3)}{1 – (1/3)^2} = \frac{2/3}{1 – 1/9} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$$

Now use the formula $\tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B}$:
$$\tan(2\theta – 2\phi) = \frac{\frac{12}{5} – \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}}$$
$$= \frac{\frac{48 – 15}{20}}{1 + \frac{36}{20}} = \frac{\frac{33}{20}}{\frac{56}{20}} = \frac{33}{56}$$

Ans. (2)