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