JEE Advanced 2022 — Paper 2 — MATH-I solution

JEE Advanced2022Paper 2MATH-I

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 _______.

Your answer

Solution

Pair the terms by common denominators:

$\frac{sin α}{cos β} + \frac{sin β}{cos α} = \frac{sin α cos α + sin β cos β}{cos α cos β}$ and $\frac{cos α}{sin β} + \frac{cos β}{sin α} = \frac{sin α cos α + sin β cos β}{sin α sin β}$ .

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

Adding gives $\frac{2}{9} (\frac{1}{cos α cos β} + \frac{1}{sin α sin β}) = \frac{2}{9} \cdot \frac{sin α sin β + cos α cos β}{sin α cos α sin β cos β} = \frac{2}{9} \cdot \frac{cos ( α - β )}{\frac{1}{4} sin 2 α sin 2 β}$ .

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

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: $(- \frac{4}{3})^{2} = \frac{16}{9} \approx 1.78$ . Therefore $⌊ \frac{16}{9} ⌋ = 1$ .

← Back to full paper

Practice more MATH-I

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →