Let α and β be real numbers such that -(π)/4<β<0<α<(π)/4. If sinα+β=1/3 and cosα-β=2/3, then the greatest integer less than or equal to sinαcosβ+cosβsinα+cosαsinβ+sinβcosα^2 is

Correct answer: 1. sinαcosβ+cosαsinβ+cosβsinα+sinβcosα^2 =cosα-βsinβcosβ+cosα-βsinαcosα^2 =2cosα-β1sin2β+1sin2α^2 =16/92sinα+β·cosα-βsin2α·sin2β^2 =16/94·13·23cos2α-2β-cos2α+2β^2 =16/9892cos2α-β-1-1-2sin2α+β^2 =16/98989-2+29^2 =16/9 i.e. 16/9=1

Q19 / 54 · JEE Advanced 2022 (Paper 2) · Mathematics

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

Correct answer: 1

Solution

${(\frac{\sin α}{\cos β} + \frac{\cos α}{\sin β} + \frac{\cos β}{\sin α} + \frac{\sin β}{\cos α})}^{2}$
$= {(\frac{\cos (α - β)}{\sin β \cos β} + \frac{\cos (α - β)}{\sin α \cos α})}^{2}$
$= {(2 \cos (α - β) \{\frac{1}{\sin 2 β} + \frac{1}{\sin 2 α}\})}^{2}$
$= \frac{16}{9} {(\frac{2 \sin (α + β) \cdot \cos (α - β)}{\sin 2 α \cdot \sin 2 β})}^{2}$
$= \frac{16}{9} {(\frac{4 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\cos (2 α - 2 β) - \cos (2 α + 2 β)})}^{2}$
$= \frac{16}{9} {(\frac{\frac{8}{9}}{(2 \cos^{2} (α - β) - 1) - (1 - 2 \sin^{2} (α + β))})}^{2}$
$= \frac{16}{9} {(\frac{\frac{8}{9}}{\frac{8}{9} - 2 + \frac{2}{9}})}^{2}$
$= \frac{16}{9}$
i.e. $[\frac{16}{9}] = 1$

Hence, the correct answer is 1.

Let α and β be real numbers such that -π4<β<0<α<π4. If sinα+β=13 and cosα-β=23, then the greatest integer less than or equal to sinαcosβ+cosβsinα+cosαsinβ+sinβcosα2 is

Solution

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