A) \[\frac{2a}{b}\]
B) \[\frac{2b}{a}\]
C) \[\frac{a}{b}\]
D) \[\frac{b}{a}\]
Correct Answer: B
Solution :
\[\tan \left[ \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} \right]+\tan \left[ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\frac{a}{b} \right]\] Let \[\frac{1}{2}{{\cos }^{-1}}\frac{a}{b}=\theta \Rightarrow \cos 2\theta =\frac{a}{b}\] Thus, \[\tan \left[ \frac{\pi }{4}+\theta \right]+\tan \left[ \frac{\pi }{4}-\theta \right]\] = \[\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=\frac{{{(1+\tan \theta )}^{2}}+{{(1-\tan \theta )}^{2}}}{(1-{{\tan }^{2}}\theta )}\] = \[\frac{1+{{\tan }^{2}}\theta +2\tan \theta +1+{{\tan }^{2}}\theta -2\tan \theta }{(1+{{\tan }^{2}}\theta )}\] = \[\frac{2(1+{{\tan }^{2}}\theta )}{1-{{\tan }^{2}}\theta }=2\sec 2\theta =\frac{2}{\cos 2\theta }\] = \[\frac{2}{a/b}=\frac{2b}{a}\].
You need to login to perform this action.
You will be redirected in
3 sec
