Let (α+β)=-(1)/(10) and (α-β)=(3)/(8) , where 0<α<(π)/(3) and 0<β<(π)/(4) . If 2 α= 3(1-r 5 ) 11 (s+ 5 ) , r, s ∈ N , then r+s is equal to \_\_\_\_ .

Correct answer: 0 2 α= [(α+β)+(α-β)] 2 α=( (α+β)+ (α-β))/(1- (α+β) · (α-β)) 2 α= (- 99 + 3 55 ) 1-( 99 )( 3 55 ) 2 α= -3 11 + 3 5 × 11 1+ 9 11 5 × 11 2 α= 3(1-11 5 ) 11 (9+ 5 ) r=11, ~s=9 r+s=20

JEE Main 202622 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Trigonometric Ratios & Identities

Let (α+β)=-(1)/(10) and (α-β)=(3)/(8) , where 0<α<(π)/(3) and...

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 N

, then

r+s

is equal to

\_\_\_\_

Correct answer

Step-by-step solution

\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})}

\mathrm{r}=11, \mathrm{~s}=9

\mathrm{r}+\mathrm{s}=20

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