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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

If

\cos x+\cos y=\;\frac{2}{3}

\sin\;x-\sin\;y=\;\frac{3}{4}

\sin\;(x-y)+\cos(x-y)=

[TG EAPCET 9 May 2024 Shift 2]

$\frac{161}{145}$
$\frac{127}{145}$
$\frac{1}{2}$
$\frac{8}{9}$

Solution:

We have,

\cos x+\cos y=\;\frac{2}{3}\;\text{ and }\;\sin\;x-\sin\;y=\;\frac{3}{4}

\therefore 2 \cos\left(\;\frac{x+y}{2}\right) \cos\left(\;\frac{x-y}{2}\right) = \frac{2}{3} \;\; \cdots

and

2 \cos\left(\;\frac{x+y}{2}\right) \sin\;\left(\;\frac{x-y}{2}\right) = \frac{3}{4}

Using Eqs. (i) and (ii), we get

\tan\left(\;\frac{x-y}{2}\right) = \frac{9}{8}

\because \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}

\therefore \tan(x-y) = \;\frac{2\left(\;\frac{9}{8}\right)}{1-\left(\;\frac{9}{8}\right)^2} = -\frac{144}{17}

\text{ Now, } \sin (x-y) = \frac{144}{145} \text{ and } \cos (x-y) = -\frac{17}{145}

\therefore \sin(x-y)+\cos(x-y) = \frac{144}{145} + \left(-\frac{17}{145}\right)

\;\; = \frac{127}{145}

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