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Question: Real part of (1 – cos q + 2i sin q)<sup>–1</sup> is –...

Real part of (1 – cos q + 2i sin q)–1 is –

A

1(3+5cos⁡θ)\frac{1}{(3 + 5\cos\theta)}

B

1(5+3cos⁡θ)\frac{1}{(5 + 3\cos\theta)}

C

1(5−3cos⁡θ)\frac{1}{(5 - 3\cos\theta)}

D

1(3−5cos⁡θ)\frac{1}{(3 - 5\cos\theta)}

Answer

1(5+3cos⁡θ)\frac{1}{(5 + 3\cos\theta)}

Explanation

Solution

Sol. 11−cos⁡θ+2isin⁡θ\frac{1}{1 - \cos\theta + 2i\sin\theta} =1−cos⁡θ−2isin⁡θ(1−cos⁡θ)2+4sin⁡2θ\frac{1 - \cos\theta - 2i\sin\theta}{(1 - \cos\theta)^{2} + 4\sin^{2}\theta}

Real part = 1−cos⁡θ(1−cos⁡θ)2+4(1−cos⁡2θ)\frac{1 - \cos\theta}{(1 - \cos\theta)^{2} + 4(1 - \cos^{2}\theta)}

= 11−cos⁡θ+4(1+cos⁡θ)\frac{1}{1 - \cos\theta + 4(1 + \cos\theta)}= 15+3cos⁡θ\frac{1}{5 + 3\cos\theta}