JEE Advanced 2022 Paper 2, Mathematics Section 1 Q1: Trigonometric Ratios And Identities

Pair the terms by common denominators:

$\frac{\sin \alpha }{\cos \beta }+\frac{\sin \beta }{\cos \alpha }=\frac{\sin \alpha \cos \alpha +\sin \beta \cos \beta }{\cos \alpha \cos \beta }$ and $\frac{\cos \alpha }{\sin \beta }+\frac{\cos \beta }{\sin \alpha }=\frac{\sin \alpha \cos \alpha +\sin \beta \cos \beta }{\sin \alpha \sin \beta }$.

Note $\sin \alpha \cos \alpha +\sin \beta \cos \beta =\frac{1}{2}(\sin 2\alpha +\sin 2\beta )=\sin (\alpha +\beta )\cos (\alpha -\beta )=\frac{1}{3}\cdot \frac{2}{3}=\frac{2}{9}$.

Adding gives $\frac{2}{9}\left(\frac{1}{\cos \alpha \cos \beta }+\frac{1}{\sin \alpha \sin \beta }\right)=\frac{2}{9}\cdot \frac{\sin \alpha \sin \beta +\cos \alpha \cos \beta }{\sin \alpha \cos \alpha \sin \beta \cos \beta }=\frac{2}{9}\cdot \frac{\cos (\alpha -\beta )}{\frac{1}{4}\sin 2\alpha \sin 2\beta }$.

Now $\sin 2\alpha \sin 2\beta =\frac{1}{2}[\cos (2\alpha -2\beta )-\cos (2\alpha +2\beta )]=\frac{1}{2}\left[(2{\cos }^2(\alpha -\beta )-1)-(1-2{\sin }^2(\alpha +\beta ))\right]=\frac{1}{2}\left[\frac{8}{9}-1-1+\frac{2}{9}\right]\cdot (-1)$. Carefully: $2\cdot \frac{4}{9}-1-1+2\cdot \frac{1}{9}=\frac{8}{9}+\frac{2}{9}-2=-\frac{8}{9}$, so $\sin 2\alpha \sin 2\beta =-\frac{4}{9}$.

Hence the sum $=\frac{2}{9}\cdot \frac{2/3}{\frac{1}{4}\cdot (-\frac{4}{9})}=\frac{2}{9}\cdot \frac{2/3}{-1/9}=-\frac{4}{3}$.

Squaring: ${\left(-\frac{4}{3}\right)}^2=\frac{16}{9}\approx 1.78$. Therefore $\lfloor \frac{16}{9}\rfloor =1$.