A) 0
B) \[\frac{2a}{b}\]
C) \[\frac{2b}{a}\]
D) \[\frac{\pi }{2}\]
Correct Answer: C
Solution :
[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 \[=\tan \left( \frac{\pi }{4}+\theta \right)+\tan \left( \frac{\pi }{4}-\theta \right)\] \[=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }\] \[=\frac{{{(1+tan\theta )}^{2}}+{{(1-tan\theta )}^{2}}}{(1-tan\theta )(1+tan\theta )}\] \[=\frac{2+2{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }=\frac{2(1+ta{{n}^{2}}\theta )}{1-{{\tan }^{2}}\theta }\] \[=\frac{2(co{{s}^{2}}\theta +si{{n}^{2}}\theta )}{(co{{s}^{2}}\theta -si{{n}^{2}}\theta )}\] \[=\frac{2}{\cos 2\theta }=\frac{2}{\frac{a}{b}}=\frac{2b}{a}\]
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