Complex Numbers And Quadratic Equations question 444
Question: If $ a=\cos \theta +i\sin \theta , $ then $ \frac{1+a}{1-a}= $ [Karnataka CET 2000]
Options:
A) $ \cot \theta $
B) $ \cot \frac{\theta }{2} $
C) $ i\cot \frac{\theta }{2} $
D) $ i\tan \frac{\theta }{2} $
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Answer:
Correct Answer: C
Solution:
$ a=\cos \theta +i\sin \theta . $ \ $ \frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }. $ Rationalization of denominator, we get $ \frac{1+a}{1-a}=\frac{(1+\cos \theta )+i\sin \theta }{(1-\cos \theta )-i\sin \theta }\times \frac{(1-\cos \theta )+i\sin \theta }{(1-\cos \theta )+i\sin \theta } $ $ =\frac{(1+\cos \theta )(1-\cos \theta )+(1+\cos \theta )i\sin \theta +(1-\cos \theta )i\sin \theta +i^{2}{{\sin }^{2}}\theta }{{{(1-\cos \theta )}^{2}}-{{(i\sin \theta )}^{2}}} $ $ =\frac{1-({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+2i\sin \theta }{1+({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )-2\cos \theta } $ $ =\frac{2i\sin \theta }{2(1-\cos \theta )} $ $ =\frac{i.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}} $ $ =i\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=i\cot \frac{\theta }{2} $ .
BETA
