Question

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

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

$=\frac{1+{\tan }^2\varphi +2\tan \varphi +1+{\tan }^2\varphi -2\tan \varphi }{1-{\tan }^2\varphi }$
$=\frac{2(1+{\tan }^2\varphi )}{1-{\tan }^2\varphi }$
$=\frac{2({\cos }^2\varphi +{\sin }^2\varphi )}{{\cos }^2\varphi -{\sin }^2\varphi }$
$=\frac{2}{\cos 2\varphi }$
$=\frac{2b}{a}$ ....... $[\because \frac{1}{2}{\cos }^{-1}\left(\frac{a}{b}\right)=\varphi \Rightarrow \cos 2\varphi =\frac{a}{b}]$