Inverse Trigonometric Functions Question 36
Question: $ {{\cos }^{-1}}( \frac{3+5\cos x}{5+3\cos x} ) $ is equal to
[Kerala (Engg.) 2005]
Options:
A) $ {{\tan }^{-1}}( \frac{1}{2}\tan \frac{x}{2} ) $
B) $ 2{{\tan }^{-1}}( 2\tan \frac{x}{2} ) $
C) $ \frac{1}{2}{{\tan }^{-1}}( 2\tan \frac{x}{2} ) $
D) $ 2{{\tan }^{-1}}( \frac{1}{2}\tan \frac{x}{2} ) $
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Answer:
Correct Answer: D
Solution:
We take $ x=\frac{\pi }{2} $ then $ \cos x=0 $
$ {{\cos }^{-1}}( \frac{3+5\cos x}{5+3\cos x} )={{\cos }^{-1}}( \frac{3}{5} ) $
$ ={{\tan }^{-1}}( \frac{4}{3} ) $ Put $ x=\frac{\pi }{2} $ in $ 2{{\tan }^{-1}}( \frac{1}{2}\tan \frac{x}{2} ) $ we get $ 2{{\tan }^{-1}}( \frac{1}{2}\tan \frac{\pi }{4} ) $
$ =2{{\tan }^{-1}}( \frac{1}{2} )={{\tan }^{-1}}( \frac{2.\frac{1}{2}}{1-\frac{1}{4}} ) $
$ ={{\tan }^{-1}}( \frac{4}{3} ) $ .
BETA
