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

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Trigonometric Ratios and Identities

Let

α

and

β

be real numbers such that

- \frac{π}{4} < β < 0 < α < \frac{π}{4}

. If

sin (α + β) = \frac{1}{3}

and

cos (α - β) = \frac{2}{3}

, then the greatest integer less than or equal to

(\frac{sin α}{cos β} + \frac{cos β}{sin α} + \frac{cos α}{sin β} + \frac{sin β}{cos α})^{2}

is ____.

Answer: 1

Given

α \in (0, \frac{π}{4})

and

β \in (- \frac{π}{4}, 0)

, so

α + β \in (- \frac{π}{4}, \frac{π}{4})

.

The expression can be rewritten as:

(\frac{cos ( α - β )}{cos β sin β} + \frac{cos ( β - α )}{sin α cos α})^{2} = 4 cos^{2} (α - β) (\frac{1}{sin 2 β} + \frac{1}{sin 2 α})^{2}

Using the identity

sin 2 α sin 2 β = \frac{1}{2} [cos 2 (α - β) - cos 2 (α + β)]

, and substituting the given values:

After simplification:

= \frac{64 \times \frac{16}{81} \times \frac{1}{9}}{( 2 \times \frac{4}{9} - 2 + \frac{2}{9} ) ^{2}} = \frac{64 \times 16}{81 \times 9} \times \frac{81}{64} = \frac{16}{9}

Therefore,

[\frac{16}{9}] = 1

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer