Trigonometric Identities Question 314
Question: If $ 90{}^\circ <A<180{}^\circ $ and $ \sin A=\frac{4}{5}, $ then $ \tan \frac{A}{2} $ is equal to
[AMU 2001]
Options:
A) $ 1/2 $
B) $ 3/5 $
C) $ 3/2 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \sin ,A=\frac{4}{5} $
Þ $ \tan A=-\frac{4}{3} $ , $ (90^{o}<A<180^{o}) $ $ \tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}} $ , (Let $ \tan \frac{A}{2}=P $ )
Þ $ -\frac{4}{3}=\frac{2P}{1-P^{2}} $
Þ $ 4P^{2}-6P-4=0 $
Þ $ P=\frac{-1}{2}\text{ (impossible),}, $ hence $ \tan \frac{A}{2}=2 $ .
BETA
