JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

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 ______.

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Solution

From $cos (α + β) = - \frac{1}{10}$ : $sin (α + β) = \frac{99}{10} = \frac{3 11}{10}$ (positive since $α + β < \frac{7 π}{12}$ ), so $tan (α + β) = - 99 = - 311$ .

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

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

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

Numerator $= - 311 \cdot 55 + 3 = - 3 \cdot 115 + 3 = 3 (1 - 115)$ .

Denominator $= 55 + 911 = 11 (5 + 9)$ .

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

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