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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If α,β∈(−π2,0)\alpha,\beta\in\left(-\dfrac{\pi}{2},0\right) such that (sin⁡α+sin⁡β)+sin⁡αsin⁡β=0(\sin\alpha+\sin\beta)+\dfrac{\sin\alpha}{\sin\beta}=0 and (sin⁡α+sin⁡β)⋅sin⁡αsin⁡β=−1(\sin\alpha+\sin\beta)\cdot\dfrac{\sin\alpha}{\sin\beta}=-1, and λ=lim⁡n→∞1+(2sin⁡α)2n(2sin⁡β)2n\lambda=\displaystyle\lim_{n\to\infty}\dfrac{1+(2\sin\alpha)^{2n}}{(2\sin\beta)^{2n}}, then
Aα=−π6\alpha=-\dfrac{\pi}{6}correct
Bλ=2\lambda=2correct
Cα=−π3\alpha=-\dfrac{\pi}{3}
Dλ=1\lambda=1
Solution
Step 1: From the two given conditions, let u=sin⁡α+sin⁡βu=\sin\alpha+\sin\beta and v=sin⁡αsin⁡βv=\dfrac{\sin\alpha}{\sin\beta}. Then
u+v=0anduv=−1u+v=0\quad\text{and}\quad uv=-1
 So v=−uv=-u and u(−u)=−1  ⇒  u2=1  ⇒  u=±1u(-u)=-1\;\Rightarrow\;u^{2}=1\;\Rightarrow\;u=\pm 1. Step 2: Since α,β∈(−π2,0)\alpha,\beta\in\left(-\dfrac{\pi}{2},0\right), both sin⁡α<0\sin\alpha<0 and sin⁡β<0\sin\beta<0, so u=sin⁡α+sin⁡β<0u=\sin\alpha+\sin\beta<0. Hence u=−1u=-1 and v=1v=1. Step 3: From v=1v=1: sin⁡α=sin⁡β\sin\alpha=\sin\beta. Combined with α,β∈(−π2,0)\alpha,\beta\in\left(-\dfrac{\pi}{2},0\right) (where sine is one-to-one), this gives α=β\alpha=\beta. Step 4: From u=−1u=-1: 2sin⁡α=−1  ⇒  sin⁡α=−12  ⇒  α=−π62\sin\alpha=-1\;\Rightarrow\;\sin\alpha=-\dfrac{1}{2}\;\Rightarrow\;\alpha=-\dfrac{\pi}{6}. Step 5: Compute λ\lambda. Since 2sin⁡α=−12\sin\alpha=-1 and 2sin⁡β=−12\sin\beta=-1:
λ=lim⁡n→∞1+(−1)2n(−1)2n=lim⁡n→∞1+11=2\lambda=\lim_{n\to\infty}\dfrac{1+(-1)^{2n}}{(-1)^{2n}}=\lim_{n\to\infty}\dfrac{1+1}{1}=2
 Correct answers: (1) α=−π6\alpha=-\dfrac{\pi}{6} and (2) λ=2\lambda=2
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