Question

Prove: \(tan^{-1}\frac 15+tan^{-1}\frac 17+tan^{-1}\frac 13+tan^{-1}\frac 18=\frac {\pi}{4}\)

Answer 1

L.H.S = tan^-1\(\frac 15\) + tan^-1\(\frac 17\) + tan^-1\(\frac 13\) + tan^-1\(\frac 18\)
          = tan^-1\((\frac {\frac 15+ \frac 17}{1-\frac 15. \frac 17})\) + tan-1\((\frac {\frac 13+ \frac 18}{1-\frac 13. \frac 18})\)       [tan^-1x+tan^-1y = tan^-1\(\frac {x+y}{1-xy}\)] 
          = tan^-1\(\frac {12}{34}\) + tan^-1\(\frac {11}{23}\) 
          = tan^-1\(\frac {6}{17}\) + tan^-1\(\frac {11}{23}\) 
          = tan^-1\((\frac {\frac {6}{17}+ \frac {11}{23}}{1-\frac {6}{17}. \frac {11}{23}})\)
          = tan^-1\(\frac {325}{325}\)
          = tan-1(1)
          = \(\frac {\pi}{4}\)
          = R.H.S