If sin x + cos y = (1)/(3) and cos x + sin y = (3)/(4) , then the value of ((x - y)/(2))is equal to

Correct answer: C. - (5)/(13) sin x + ((π)/(2) - y) = (1)/(3) ⇒ 2 ((π)/(4) + (x - y)/(2)) ((x + y)/(2) - (π)/(4)) = (1)/(3) cos x + ((π)/(2) - y) = (3)/(4) ⇒ 2 ((π)/(4) + (x - y)/(2)) ((x + y)/(2) - (π)/(4)) = (3)/(4) On dividing we get, ((π)/(4) + (x - y)/(2)) = (4)/(9) ⇒ (1 + ((x - y)/(2)))/(1 - ((x - y)/(2))) = (4)/(9) Using componendo and dividendo, we get (2)/(2 ((x - y)/(2))) = (13)/(- 5) ⇒ ((x - y)/(2)) = - (5)/(13)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If sin x + cos y = (1)/(3) and cos x + sin y = (3)/(4) , then the...

sin x + cos y = \dfrac{1}{3}

and

cos x + sin y = \dfrac{3}{4}

, then the value of

\tan \left(\dfrac{x - y}{2}\right)is equal to

Options

A $\dfrac{5}{13}$
B $\dfrac{12}{13}$
C $- \dfrac{5}{13}$
D $- \dfrac{5}{12}$

Correct answer

C. $- \dfrac{5}{13}$

Step-by-step solution

sin x + \sin \left(\dfrac{\pi}{2} - y\right) = \dfrac{1}{3}

⇒ 2\sin \left(\dfrac{\pi}{4} + \dfrac{x - y}{2}\right)\cos \left(\dfrac{x + y}{2} - \dfrac{\pi}{4}\right) = \dfrac{1}{3}

cos x + \cos \left(\dfrac{\pi}{2} - y\right) = \dfrac{3}{4}

⇒ 2\cos \left(\dfrac{\pi}{4} + \dfrac{x - y}{2}\right)\cos \left(\dfrac{x + y}{2} - \dfrac{\pi}{4}\right) = \dfrac{3}{4}

On dividing we get,

\tan \left(\dfrac{\pi}{4} + \dfrac{x - y}{2}\right) = \dfrac{4}{9}

⇒ \dfrac{1 + \tan \left(\dfrac{x - y}{2}\right)}{1 - \tan \left(\dfrac{x - y}{2}\right)} = \dfrac{4}{9}

Using componendo and dividendo, we get

\dfrac{2}{2\tan \left(\dfrac{x - y}{2}\right)} = \dfrac{13}{- 5} ⇒ \tan \left(\dfrac{x - y}{2}\right) = - \dfrac{5}{13}

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