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 ______。

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Trigonometric Ratios and Identities

Let

cos (α + β) = - \frac{1}{10}

and

sin (α - β) = \frac{3}{8}

, where

0 < α < \frac{π}{3}

and

0 < β < \frac{π}{4}

.

If

tan 2 α = \frac{3 ( 1 - r 5 )}{11 ( s + 5 )}, r, s \in N

, then

r + s

is equal to ______。

Answer: 20

Write

2 α = (α + β) + (α - β)

. With

cos (α + β) = - \frac{1}{10}

and the given ranges placing

α + β

in the second quadrant,

sin (α + β) = \frac{3 11}{10}

, so

tan (α + β) = - 311

. From

sin (α - β) = \frac{3}{8}

cos (α - β) = \frac{55}{8}

, so

tan (α - β) = \frac{3}{55}

. Then

tan 2 α = \frac{- 3 11 + \frac{3}{55}}{1 + \frac{9 11}{55}} = \frac{3 ( 1 - 11 5 )}{11 ( 9 + 5 )},

after multiplying numerator and denominator by

55

. Comparing with the given form,

r = 11

and

s = 9

, so

r + s = 20

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer