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MHT CET 2021 Math Solved Paper

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Question : 45 of 100

Marks: +1, -0

If

\; \frac{3}{2+\cos\theta + i\sin\theta} = a + i b

[(a-2)^{2}+b^{2}]

0
1
$-1$
2

Solution:

Given,

\; \frac{3}{2+\cos\theta + i\sin\theta} = a + i b

\Rightarrow \;\; \frac{3[(2+\cos\theta) - i\sin\theta]}{(2+\cos\theta)^{2} + \sin^{2}\theta} = a + i b

\Rightarrow \;\; \frac{3[2+\cos\theta - i\sin\theta]}{5+4\cos\theta} = a + i b

\Rightarrow \;\; a = \; \frac{3(2+\cos\theta)}{5+4\cos\theta}

and

\;\; b = - \; \frac{3\sin\theta}{5+4\cos\theta}

\therefore (a-2)^{2} + b^{2} = \left( \frac{6+3\cos\theta}{5+4\cos\theta} - 2 \right)^{2} + \frac{9\sin^{2}\theta}{(5+4\cos\theta)^{2}}

= \frac{(-4-5\cos\theta)^{2} + 9\sin^{2}\theta}{(5+4\cos\theta)^{2}}

= \frac{16+25\cos^{2}\theta + 40\cos\theta + 9\sin^{2}\theta}{(5+4\cos\theta)^{2}}

= \frac{16+16\cos^{2}\theta + 40\cos\theta + 9}{(5+4\cos\theta)^{2}}

= \frac{(5+4\cos\theta)^{2}}{(5+4\cos\theta)^{2}} = 1

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