If 0 < x, y < π and x + y - (x + y) = (3)/(2) , then x + y is equal to:

Correct answer: D. 1 + 3 2 x + y - (x + y) = (3)/(2) 2 ((x + y)/(2)) · ((x - y)/(2)) - [2 ^ 2 ((x + y)/(2)) - 1] = (3)/(2) 2 ((x + y)/(2)) · ((x - y)/(2)) - 2 ^ 2 ((x + y)/(2)) + 1 = (3)/(2) 2 ((x + y)/(2)) · ((x - y)/(2)) - 2 ^ 2 ((x + y)/(2)) = (1)/(2) 4 ((x + y)/(2)) · ((x - y)/(2)) - 4 ^ 2 ((x + y)/(2)) = 1 ⇒ 4 ((x + y)/(2)) · ((x - y)/(2)) - 4 ^ 2 ((x + y)/(2)) = 1 = ^ 2 ((x - y)/(2)) + ^ 2 ((x - y)/(2)) ⇒ 4 ^ 2 ((x + y)/(2)) + ^ 2 ((x - y)/(2)) - 4 ((x + y)/(2)) · ((x - y)/(2)) + ^ 2 ((x - y)/(2)) = 0 ⇒ (2 ((x + y)/(2)) - ((x - y)/(2))) ^ 2 + ^ 2 ((x - y)/(2)) = 0 ⇒ ((x - y)/(2)) = 0 and ((x + y)/(2)) = (1)/(2) ((x - y)/(2)) ⇒ x = y and x = (1)/(2) = y ∴ x = 3 2 ⇒ x + y = 1 + 3 2 option (4)

JEE Main 202125 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 25 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Equations

JEE Main 2021 · 25 February · Evening Shift · Mathematics · Trigonometric Equations

If 0 < x, y < π and x + y - (x + y) = (3)/(2) , then x + y is equal...

0 < x, y < \pi

and

\cos x + \cos y - \cos \left(x + y\right) = \dfrac{3}{2}

, then

\sin x + \cos y

is equal to:

Options

A $\dfrac{1}{2}$
B $\dfrac{\sqrt{3}}{2}$
C $\dfrac{1 - \sqrt{3}}{2}$
D $\dfrac{1 + \sqrt{3}}{2}$

Correct answer

D. $\dfrac{1 + \sqrt{3}}{2}$

Step-by-step solution

\cos x + \cos y - \cos \left(x + y\right) = \dfrac{3}{2} $$ 2\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) - \left[2\cos^{2}\left(\dfrac{x + y}{2}\right) - 1\right] = \dfrac{3}{2} $$ 2\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) - 2\cos^{2}\left(\dfrac{x + y}{2}\right) + 1 = \dfrac{3}{2} $$ 2\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) - 2\cos^{2}\left(\dfrac{x + y}{2}\right) = \dfrac{1}{2} $$ 4\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) - 4\cos^{2}\left(\dfrac{x + y}{2}\right) = 1 $$ ⇒ 4\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) - 4\cos^{2}\left(\dfrac{x + y}{2}\right) = 1 = \sin^{2}\left(\dfrac{x - y}{2}\right) + \cos^{2}\left(\dfrac{x - y}{2}\right) $$ ⇒ 4\cos^{2}\left(\dfrac{x + y}{2}\right) + \cos^{2}\left(\dfrac{x - y}{2}\right) - 4\cos \left(\dfrac{x + y}{2}\right) \cdot \cos \left(\dfrac{x - y}{2}\right) + \sin^{2}\left(\dfrac{x - y}{2}\right) = 0 $$ ⇒ {\left(2\cos \left(\dfrac{x + y}{2}\right) - \cos \left(\dfrac{x - y}{2}\right)\right)}^{2} + \sin^{2}\left(\dfrac{x - y}{2}\right) = 0 $$ ⇒ \sin \left(\dfrac{x - y}{2}\right) = 0 $and$ \cos \left(\dfrac{x + y}{2}\right) = \dfrac{1}{2}\cos \left(\dfrac{x - y}{2}\right) $⇒ x = y and$ \cos x = \dfrac{1}{2} = \cos y $$ ∴ \sin x = \dfrac{\sqrt{3}}{2} $$ ⇒ \sin x + \cos y = \dfrac{1 + \sqrt{3}}{2} $option \left(4\right)$

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