Question

The value of $tan({\cos }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{1}{4})$ is

• A. $\frac{19}{8}$
• B. $\frac{8}{19}$
• C. $\frac{19}{12}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $\frac{19}{8}$

Given, $\tan ({\cos }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{1}{4})$

Let, ${\cos }^{-1}\frac{3}{5}=a$

$\Rightarrow \cos a=\frac{3}{5}$

$\Rightarrow {\sin }^{2}a=1-{\cos }^{2}a=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$

$\therefore \sin a=\frac{4}{5}$

$\Rightarrow \tan a=\frac{\sin a}{\cos a}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$

$\Rightarrow a={\tan }^{-1}\frac{4}{3}$

$\therefore \tan ({\cos }^{-1}\frac{3}{5}+{\tan }^{-1}\frac{1}{4})$

$=\tan (a+{\tan }^{-1}\frac{1}{4})$ where $a={\cos }^{-1}\frac{3}{5}$

$=\tan ({\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{4})$ where $a={\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{4}{3}$

We know that ${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

$=\tan \left({\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{4}{3}\times \frac{1}{4}}\right)\right)$

$=\tan \left({\tan }^{-1}\left(\frac{\frac{16+3}{12}}{1-\frac{4}{12}}\right)\right)$

$=\tan \left({\tan }^{-1}\left(\frac{\frac{19}{12}}{\frac{12-4}{12}}\right)\right)$

$=\tan \left({\tan }^{-1}\frac{19}{8}\right)$

$=\frac{19}{8}$