Trigonometric Identities Question 113
Question: If $ \tan A=\frac{1-\cos B}{\sin B}, $ find $ \tan 2A $ in terms of $ \tan B $ and show that
[IIT 1983; MP PET 1994]
Options:
A) $ \tan 2A=\tan B $
B) $ \tan 2A={{\tan }^{2}}B $
C) $ \tan 2A={{\tan }^{2}}B+2\tan B $
D) None of the above
Show Answer
Answer:
Correct Answer: A
Solution:
$ \tan A=\frac{1-\cos B}{\sin B}=\frac{2{{\sin }^{2}}(B/2)}{2\sin (B/2)\cos (B/2)}=\tan \frac{B}{2} $
Þ $ \tan 2A=\tan B $ .
BETA
