Inverse Trigonometric Functions Question 172
Question: If $ {{\sin }^{-1}}( \frac{2a}{1+a^{2}} )+{{\sin }^{-1}}( \frac{2b}{1+b^{2}} )=2{{\tan }^{-1}}x, $ then $ x= $
[MNR 1984; UPSEAT 1999; Pb. CET 2004]
Options:
A) $ \frac{a-b}{1+ab} $
B) $ \frac{b}{1+ab} $
C) $ \frac{b}{1-ab} $
D) $ \frac{a+b}{1-ab} $
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Answer:
Correct Answer: D
Solution:
$ {{\sin }^{-1}}( \frac{2a}{1+a^{2}} )+{{\sin }^{-1}}( \frac{2b}{1+b^{2}} )=2{{\tan }^{-1}}x $ Putting $ a=\tan \theta $ and $ b=\tan \varphi $ .
So, $ {{\sin }^{-1}}( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } )+{{\sin }^{-1}}( \frac{2\tan \varphi }{1+{{\tan }^{2}}\varphi } )=2{{\tan }^{-1}}x $
Therefore $ {{\sin }^{-1}}\sin (2\theta )+{{\sin }^{-1}}\sin (2\varphi )=2{{\tan }^{-1}}x $
Therefore $ 2(\theta +\varphi )=2{{\tan }^{-1}}x $ Hence $ x=\tan (\theta +\varphi ) $
Therefore $ x=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi } $ Substituting these values, we get $ x=\frac{a+b}{1-ab} $ .
BETA
