Prove that :`2tan^(-1)1/5+sec^(-1)(5sqrt(2))/7+2tan^(-1)\ 1/8=pi/4`

Explanation

Let `sec^(-1)(5sqrt(2))/7=theta`. Then, `sec theta=(5sqrt(2))/7`. So, `cos theta=7/(5sqrt(2))`.
Draw a right triangle with base = 7 units and hypotenuse =`5sqrt(2)` units.
`therefore` Perpendicular `=sqrt((5sqrt(2))^(2)-7^(2))=sqrt(50-49)=1` unit.
`therefore tantheta=1/7 rArr theta=tan^(-1)1/7`.
`therefore` LHS =`2(tan^(-1)1/5+tan^(-1)1/8)+tan^(-1)1/7`.
`=2tan^(-1)(13/40 xx 40/30)+tan^(-1)1/7 = 2tan^(-1)1/3+tan^(-1)1/7`
`=tan^(-1){(2 xx 1/3)/(1-(1/3)^(2)}+tan^(-1)1/7=2tan^(-1)1/3+tan^(-1)1/7`
`=tan^(-1)((3/4+1/7))/(1-3/4 xx 1/7))= tan^(-1)25/25=tan^(-1)1`
`=pi/4`= RHS.
Hence, LHS=RHS