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${(1 - \cos \theta + 2i\sin \theta )^{ - 1}}$ का वास्तविक भाग है
$\frac{1}{{3 + 5\cos \theta }}$
$\frac{1}{{5 - 3\cos \theta }}$
$\frac{1}{{3 - 5\cos \theta }}$
$\frac{1}{{5 + 3\cos \theta }}$
Solution
(d) ${\{ (1 – \cos \theta ) + i.2\sin \theta \} ^{ – 1}} = {\left\{ {2{{\sin }^2}\frac{\theta }{2} + i.4\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right\}^{ – 1}}$
= ${\left( {2\sin \frac{\theta }{2}} \right)^{ – 1}}{\left\{ {\sin \frac{\theta }{2} + i.2\cos \frac{\theta }{2}} \right\}^{ – 1}}$
$ = {\left( {2\sin \frac{\theta }{2}} \right)^{ – 1}}\frac{1}{{\sin \frac{\theta }{2} + i.2\cos \frac{\theta }{2}}} \times \frac{{\sin \frac{\theta }{2} – i.2\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2} – i.2\cos \frac{\theta }{2}}}$
$ = \frac{{\sin \frac{\theta }{2} – i.2\cos \frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\left( {{{\sin }^2}\frac{\theta }{2} + 4{{\cos }^2}\frac{\theta }{2}} \right)}}$.
यह वास्तविक भाग है।
$ = \frac{{\sin \frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\left( {1 + 3{{\cos }^2}\frac{\theta }{2}} \right)}} = \frac{1}{{2\left( {1 + 3{{\cos }^2}\frac{\theta }{2}} \right)}}$$ = \frac{1}{{5 + 3\cos \theta }}$
