Complex Numbers And Quadratic Equations question 418
Question: The real part of $ {{(1-\cos \theta +2i\sin \theta )}^{-1}} $ is [IIT 1978, 86]
Options:
A) $ \frac{1}{3+5\cos \theta } $
B) $ \frac{1}{5-3\cos \theta } $
C) $ \frac{1}{3-5\cos \theta } $
D) $ \frac{1}{5+3\cos \theta } $
Show Answer
Answer:
Correct Answer: D
Solution:
$ {{{(1-\cos \theta )+i.2\sin \theta }}^{-1}}={{{ 2{{\sin }^{2}}\frac{\theta }{2}+i.4\sin \frac{\theta }{2}\cos \frac{\theta }{2} }}^{-1}} $ = $ {{( 2\sin \frac{\theta }{2} )}^{-1}}{{{ \sin \frac{\theta }{2}+i.2\cos \frac{\theta }{2} }}^{-1}} $ $ ={{( 2\sin \frac{\theta }{2} )}^{-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}( {{\sin }^{2}}\frac{\theta }{2}+4{{\cos }^{2}}\frac{\theta }{2} )} $ . it?s real part $ =\frac{\sin \frac{\theta }{2}}{2\sin \frac{\theta }{2}( 1+3{{\cos }^{2}}\frac{\theta }{2} )}=\frac{1}{2( 1+3{{\cos }^{2}}\frac{\theta }{2} )} $ $ =\frac{1}{5+3\cos \theta } $
BETA
