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

Explanation

Let `1/2cos^-1(a/b) = x`

Then, `a/b = cos2x->(1)`

Now,

`L.H.S. = tan(pi/4+1/2cos^-1(a/b))+tan(pi/4-1/2cos^-1(a/b))`

`=tan(pi/4+x)+tan(pi/4-x)`

`=(1+tanx)/(1-tanx)+(1-tanx)/(1+tanx)`

`=((1+tanx)^2+(1-tanx)^2)/(1-tan^2x)`

`=(1+tan^2x+2tanx+1+tan^2x-2tanx)/(1-tan^2x)`

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