Answer 1

Let `cos^(-1) (a/b) = 0`

Then `cos theta = a/b`

L.H.S:

`tan {pi/4 + 1/4 cos^(-1)  a/b} + tan (pi/4 - 1/2cos^(-1)  a/b)`

= `tan (pi/4 + theta/2) + tan (pi/4 - theta/2)`

`= (1 + tan  theta/2)/(1 - tan  theta/2) + (1 - tan  theta/2)/(1 + tan  theta/2)`

`((1 + tan  theta/2)^2 + (1 - tan  theta/2)^2)/(1 - tan^2   theta/2)`

= `2((1 + tan^2  theta/2)/(1 -  tan^2  theta/2))`

= `2/(cos theta)`   `[∵ cos 2 theta = (1 - tan^2 theta)/(1 + tan^2 theta)]`

= `(2b)/a`

=RHS

LHS = RHS

Hence Proved