Let cos (α+β)=-1/10 and sin (α-β)=3/8 , where 0<α<π/3 and 0<β<π/4 . If tan 2 α=frac 3(1-r √(5)) √(11)(s+√(5)) , r, s in mathbb N , then r+s is equal to \ \ \ \ 。

Mathematics · Trigonometry Ratios and Identities

JEE Main 2026 — 22 January, Evening Shift — Question 23

Let $\cos (\alpha+\beta)=-\frac{1}{10}$ and $\sin (\alpha-\beta)=\frac{3}{8}$ , where $0<\alpha<\frac{\pi}{3}$ and $0<\beta<\frac{\pi}{4}$ .

If $\tan 2 \alpha=\frac{3(1-r \sqrt{5})}{\sqrt{11}(s+\sqrt{5})}, r, s \in \mathbb{N}$ , then $r+s$ is equal to $\_\_\_\_$ 。

Numerical answer — enter this value.

$\tan 2 \alpha=\tan [(\alpha+\beta)+(\alpha-\beta)]$

$\tan 2 \alpha=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \cdot \tan (\alpha-\beta)}$

$\tan 2 \alpha=\frac{\left(-\sqrt{99}+\frac{3}{\sqrt{55}}\right)}{1-(\sqrt{99})\left(\frac{3}{\sqrt{55}}\right)}$

$\tan 2 \alpha=\frac{-3 \sqrt{11}+\frac{3}{\sqrt{5} \times \sqrt{11}}}{1+\frac{9 \sqrt{11}}{\sqrt{5} \times \sqrt{11}}}$

$\tan 2 \alpha=\frac{3(1-11 \sqrt{5})}{\sqrt{11}(9+\sqrt{5})}$

$r=11, s=9$

$r+s=20$

Answer key and solution verified before publishing.

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