Question

Prove That ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{4}$

Answer 1

Solve by applying trigonometric formulas:

We have ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)=\frac{\mathrm{\pi }}{8}$

Taking LHS ${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)$

As we know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

So, ${\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{8}}}{1-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{8}}}\right)+{\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{1}{5}}\times {\displaystyle \frac{1}{7}}}\right)$

$={\tan }^{-1}\frac{{\displaystyle \frac{11}{24}}}{{\displaystyle \frac{23}{24}}}+{\tan }^{-1}\frac{{\displaystyle \frac{12}{35}}}{{\displaystyle \frac{34}{35}}}$

$={\tan }^{-1}\frac{11}{23}+{\tan }^{-1}\frac{6}{17}$

Again by using ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Then, ${\tan }^{-1}\left(\frac{{\displaystyle \frac{11}{23}}+{\displaystyle \frac{6}{17}}}{1-{\displaystyle \frac{11}{23}}\times {\displaystyle \frac{6}{17}}}\right)$

$$\begin{gathered} ={\tan }^{-1}\frac{{\displaystyle \frac{138+187}{391}}}{{\displaystyle \frac{391-65}{391}}} \\ ={\tan }^{-1}\frac{{\displaystyle \frac{325}{391}}}{{\displaystyle \frac{325}{391}}} \\ ={\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left(\tan \frac{\pi }{4}\right) \\ =\frac{\pi }{4} \end{gathered}$$

Hence, proved.