Prove the following: `tan[pi/4+1/2cos^(-1)(a/b)]+tan[pi/4-1/2cos^(-1)(a/b)]=(2b)/a`

Explanation

Let `1/2​cos^(−1)a/b=θ`

`=>cos2θ=a/b`

LHS`tan(π/4​+θ)+tan(π/4​−θ)`

`=(1+tanθ​)/(1−tanθ)+(1−tanθ​)/(1+tanθ)`

`=((1+tanθ)^2+(1−tanθ)^2​)/(1−tan2θ)`

`=(2(sin^2θ+cos^2θ)​)/(cos^2θ−sin^2θ)`

`=2/(cos2θ)​=(2b)/a​`