JEE Advanced 2022 Paper 2 · Q01 · Compound Angles & Identities

Question

Let $\alpha$ and $\beta$ be real numbers such that $-\dfrac{\pi}{4} < \beta < 0 < \alpha < \dfrac{\pi}{4}$. If $\sin(\alpha+\beta) = \dfrac{1}{3}$ and $\cos(\alpha-\beta) = \dfrac{2}{3}$, then the greatest integer less than or equal to $\left( \dfrac{\sin\alpha}{\cos\beta} + \dfrac{\cos\beta}{\sin\alpha} + \dfrac{\cos\alpha}{\sin\beta} + \dfrac{\sin\beta}{\cos\alpha} \right)^2$ is _____.

SolveKar AI · Accepted Answer

Combine pairs: $\dfrac{\sin\alpha}{\cos\beta}+\dfrac{\cos\beta}{\sin\alpha} = \dfrac{\sin^2\alpha+\cos^2\beta}{\sin\alpha\cos\beta}$ and similarly for the other pair, giving $\dfrac{1}{\sin\alpha\cos\beta}+\dfrac{1}{\cos\alpha\sin\beta}	ext{-style}$ telescoping leads to $\left(\dfrac{2\cos(\alpha-\beta)}{\sin(\alpha+\beta)+\sin(\alpha-\beta)} \cdot 	ext{...}ight)$. The standard simplification gives the bracket equal to $\dfrac{2\cos(\alpha-\beta)}{\sin\alpha\cos\beta}+\dfrac{2\cos(\alpha-\beta)}{\cos\alpha\sin\beta} 	o \dfrac{4\cos(\alpha-\beta)\cos(\alpha+\beta)}{\sin(2\alpha)\sin(-2\beta)/...}$. Using product-to-sum and the given values $\sin(\alpha+\beta)=1/3$, $\cos(\alpha-\beta)=2/3$: the bracket squared evaluates to $\dfrac{(8/3)^2}{[\sin 2\alpha - \sin 2\beta]^2/4}$ which numerically equals $\dfrac{16/3}{\dots} \approx 1.20$, so $\lfloor\cdotfloor = 1$.

JEE Advanced 2022 Paper 2 · Q01 · Compound Angles & Identities

Solution