Question

If $2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{m}{7}$.Find $m$

Answer 1

As we know that,

$2{\tan }^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}$

$\therefore 2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{m}{7}$

${\tan }^{-1}\left(\frac{2\times \frac{1}{5}}{1-{\left(\frac{1}{5}\right)}^2}\right)+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{m}{7}$

$\Rightarrow {\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{m}{7}$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12}\times \frac{1}{8}}\right)={\tan }^{-1}\frac{m}{7}$

$\Rightarrow \frac{\left(\frac{10+3}{24}\right)}{\frac{96-5}{96}}=\frac{m}{7}$

$\Rightarrow \frac{4}{7}=\frac{m}{7}$

$\Rightarrow m=4$