Question

The value of $3ta{n}^{-1}\left(\frac{1}{2}\right)+2ta{n}^{-1}\left(\frac{1}{5}\right)$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. none of these

Answer 1

The correct option is A $\frac{\pi }{4}$

$3{\tan }^{-1}\frac{1}{2}+2{\tan }^{-1}\frac{1}{5}$

$2{\tan }^{-1}\frac{1}{2}={\tan }^{-1}\left(\frac{\frac{2}{2}}{1-\frac{1}{4}}\right)={\tan }^{-1}\frac{4}{3}$

$3{\tan }^{-1}\frac{1}{2}={\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{2}$

$={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3}\frac{1}{2}}\right)$

$={\tan }^{-1}\frac{11}{2}$

$3ta{n}^{-1}\frac{1}{2}+2{\tan }^{-1}\frac{1}{5}={\tan }^{-1}\frac{11}{2}+{\tan }^{-1}\frac{5}{12}$

$={\tan }^{-1}\left(\frac{\frac{11}{2}+\frac{5}{12}}{1-\frac{11}{2}\frac{5}{12}}\right)$

$={\tan }^{-1}\left(\frac{\frac{71}{12}}{\frac{-31}{24}}\right)$

$=\pi +{\tan }^{-1}(-\frac{142}{31})$