Inverse Trigonometric Functions Question 10
Question: $ \tan [ \frac{1}{2}{{\sin }^{-1}}( \frac{2a}{1+a^{2}} )+\frac{1}{2}{{\cos }^{-1}}( \frac{1-a^{2}}{1+a^{2}} ) ]= $
Options:
A) $ \frac{2a}{1+a^{2}} $
B) $ \frac{1-a^{2}}{1+a^{2}} $
C) $ \frac{2a}{1-a^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tan [ \frac{1}{2}{{\sin }^{-1}}( \frac{2a}{1+a^{2}} )+\frac{1}{2}{{\cos }^{-1}}( \frac{1-a^{2}}{1+a^{2}} ) ] $
$ =\tan [ \frac{1}{2}{{\sin }^{-1}}( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } )+\frac{1}{2}{{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) ] $
(Let $ a=\tan \theta $ ) $ =\tan [ \frac{1}{2}{{\sin }^{-1}}(\sin 2\theta )+\frac{1}{2}{{\cos }^{-1}}(\cos 2\theta ) ] $
$ =\tan (2\theta )=\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\frac{2a}{1-a^{2}} $
Trick: Put $ a=0 $ , then tan (0+0) = 0; which is given by (a) and (c).
Again put $ a=1 $ , then $ \tan ( \frac{\pi }{4}+\frac{\pi }{4} )=\infty $ , which is given by (c).
BETA
