Let cos = - 1 10 and sin = 3 8 , where 0 3 and 0 4 , If tan 2 = 3 ( 1 - r 5 ) 11 ( s + 5 ) , r , s , then r + s is equal to______. [2026]

Question

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 $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]

Smart Achievers · Accepted Answer

(20)

$\tan 2 α = \tan [(α + β) + (α - β)]$

$\tan 2 α = \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}$

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

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

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

$r = 11, s = 9$

$r + s = 20$

Solution

Q.
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 $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]

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Solution

Q. Let cos(α+β)=-110 and sin(α-β)=38, where 0<α<π3 and 0<β<π4, If tan2α=3(1-r5)11(s+5), r,s∈ℕ, then r + s is equal to______. [2026]

Ans. (20) tan2α=tan[(α+β)+(α-β)] tan2α=tan(α+β)+tan(α-β)1-tan(α+β)tan(α-β) tan2α=-99+3551-(99)(355) tan2α=-311+35×111+9115×11 tan2α=3(1-115)11(9+5) r=11, s=9 r+s=20

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Q.
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 $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]