Question

$\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]+\tan \left[\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]$ is equal to.

• A. $\frac{2a}{b}$
• B. $\frac{2b}{a}$
• C. $\frac{a}{b}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B $\frac{2b}{a}$

$\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]+\tan \left(\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right)$

Let $\frac{1}{2}{\cos }^{-1}\frac{a}{b}=\theta$

$\Rightarrow \cos 2\theta =\frac{a}{b}$

Thus, $\tan \left(\frac{\pi }{4}+\theta \right)+\tan \left(\frac{\pi }{4}-\theta \right)$

$=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }$

$=\frac{(1+\tan \theta {)}^2+(1-\tan \theta {)}^2}{(1-{\tan }^2\theta )}$

$=\frac{1+{\tan }^2\theta +2\tan \theta +1+{\tan }^2\theta -2\tan \theta }{(1-{\tan }^2\theta )}$

$=\frac{2(1+{\tan }^2\theta )}{(1-{\tan }^2\theta )}=2\sec 2\theta =\frac{2}{\cos 2\theta }$

$=\frac{2}{a/b}=\frac{2b}{a}$.