Inverse Trigonometric Functions Question 50
Question: Simplified form of $ \tan ( \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} )+\tan ( \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} ) $ is
Options:
A) 0
B) $ \frac{2a}{b} $
C) $ \frac{2b}{a} $
D) $ \frac{\pi }{2} $
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Answer:
Correct Answer: C
Let $ \frac{1}{2}{{\cos }^{-1}}\frac{a}{b}=\theta ; $
then $ {{\cos }^{-1}}\frac{a}{b}=2\theta ;\Rightarrow \cos 2\theta =\frac{a}{b} $ then expression $ =\tan ( \frac{\pi }{4}+\theta )+\tan ( \frac{\pi }{4}-\theta ) $
$ =\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta } $
$ =\frac{{{(1+tan\theta )}^{2}}+{{(1-tan\theta )}^{2}}}{(1-tan\theta )(1+tan\theta )} $
$ =\frac{2+2{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }=\frac{2(1+tan^{2}\theta )}{1-{{\tan }^{2}}\theta } $
$ =\frac{2(cos^{2}\theta +sin^{2}\theta )}{(cos^{2}\theta -sin^{2}\theta )} $
$ =\frac{2}{\cos 2\theta }=\frac{2}{\frac{a}{b}}=\frac{2b}{a} $
BETA
