JEE Main 2026 Jan 22 Shift 2, Mathematics Q23: Trigonometric Ratios And Identities

From $\cos (\alpha +\beta )=-\frac{1}{10}$: $\sin (\alpha +\beta )=\frac{\sqrt{99}}{10}=\frac{3\sqrt{11}}{10}$ (positive since $\alpha +\beta <\frac{7\pi }{12}$), so $\tan (\alpha +\beta )=-\sqrt{99}=-3\sqrt{11}$.

From $\sin (\alpha -\beta )=\frac{3}{8}$: $\cos (\alpha -\beta )=\frac{\sqrt{55}}{8}$, so $\tan (\alpha -\beta )=\frac{3}{\sqrt{55}}$.

$\tan 2\alpha =\tan [(\alpha +\beta )+(\alpha -\beta )]=\frac{-3\sqrt{11}+\frac{3}{\sqrt{55}}}{1-(-3\sqrt{11})\cdot \frac{3}{\sqrt{55}}}=\frac{-3\sqrt{11}+\frac{3}{\sqrt{55}}}{1+\frac{9\sqrt{11}}{\sqrt{55}}}$.

Multiply numerator and denominator by $\sqrt{55}=\sqrt{5}\cdot \sqrt{11}$:

Numerator $=-3\sqrt{11}\cdot \sqrt{55}+3=-3\cdot 11\sqrt{5}+3=3(1-11\sqrt{5})$.

Denominator $=\sqrt{55}+9\sqrt{11}=\sqrt{11}(\sqrt{5}+9)$.

So $\tan 2\alpha =\frac{3(1-11\sqrt{5})}{\sqrt{11}(9+\sqrt{5})}$, giving $r=11$, $s=9$, $r+s=20$.