A) \[\frac{2a}{b}\]
B) \[\frac{2b}{a}\]
C) \[\frac{a}{b}\]
D) \[\frac{b}{a}\]
Correct Answer: B
Solution :
\[\text{tan}\left[ \frac{\pi }{4}+\frac{1}{2}{{\cos }^{-1}}\left( \frac{a}{b} \right) \right]+\tan \left[ \frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}\left( \frac{a}{b} \right) \right]\] \[=\tan \left[ \frac{\pi }{4}+\phi \right]+\tan \left[ \frac{\pi }{4}-\phi \right]\] \[\left[ \text{put }\frac{1}{2}{{\cos }^{-1}}\left( \frac{a}{b} \right)=\phi \right]\] \[=\frac{1+\tan \phi }{1-\tan \phi }+\frac{1-\tan \phi }{1+\tan \phi }\] \[=\frac{{{(1+\tan \phi )}^{2}}+{{(1-\tan \phi )}^{2}}}{1-{{\tan }^{2}}\phi }\] \[=\frac{2(1+{{\tan }^{2}}\phi )}{1-{{\tan }^{2}}\phi }\] \[=\frac{2({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )}{{{\cos }^{2}}\phi -{{\sin }^{2}}\phi }\] \[=\frac{2}{\cos 2\phi }\] \[=\frac{2b}{a}\left[ \because \frac{1}{2}{{\cos }^{-1}}\left( \frac{a}{b} \right)=\phi \Rightarrow \cos 2\phi =\frac{a}{b} \right]\]
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