Question

$ta{n}^{-1}\frac{1}{5}+ta{n}^{-1}\frac{1}{7}+ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{8}=\frac{\pi }{4}$

Answer 1

$LHS=(ta{n}^{-1}\frac{1}{5}+ta{n}^{-1}\frac{1}{7})+(ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{8})ta{n}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}\times \frac{1}{7}}\right)+ta{n}^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3}\times \frac{1}{8}}\right)[\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)]=ta{n}^{-1}\left(\frac{\frac{7+5}{35}}{\frac{35-1}{35}}\right)+ta{n}^{-1}\left(\frac{\frac{8+3}{24}}{\frac{24-1}{24}}\right)=ta{n}^{-1}\left(\frac{6}{17}\right)+ta{n}^{-1}\left(\frac{11}{23}\right)=ta{n}^{-1}\left(\frac{\frac{16}{17}+\frac{11}{23}}{1-\frac{6}{17}\times \frac{11}{23}}\right)+ta{n}^{-1}\left(\frac{6\times 23+11\times 17}{17\times 23-6\times 11}\right)=ta{n}^{-1}\left(\frac{325}{325}\right)=ta{n}^{-1}1=\frac{\pi }{4}=RHS.$