Complex Numbers And Quadratic Equations question 447
Question: The real part of $ \frac{1}{1-\cos \theta +i\sin \theta } $ is equal to [Karnataka CET 2001, 05]
Options:
A) 1/4
B) 1/2
C) tan q/2
D) 1/1- cos q
Show Answer
Answer:
Correct Answer: B
Solution:
$ \frac{1}{1-\cos \theta +i\sin \theta } $ $ =\frac{1}{(1-\cos \theta )+i\sin \theta }\times \frac{(1-\cos \theta )-i\sin \theta }{(1-\cos \theta )-i\sin \theta } $ $ =\frac{(1-\cos \theta )-i\sin \theta }{{{(1-\cos \theta )}^{2}}+{{\sin }^{2}}\theta } $ $ =\frac{(1-\cos \theta )-i\sin \theta }{2(1-\cos \theta )} $ $ =\frac{(1-\cos \theta )}{2(1-\cos \theta )}-i\frac{\sin \theta }{2(1-\cos \theta )}. $ Therefore its real part = $ \frac{1-\cos \theta }{2(1-\cos \theta )}=\frac{1}{2} $
BETA
