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Prove that Tan {Pi/4 + 1by2 Cos^(-1) Abyb} + Tan {Piby4 - 1by2 Cos^(-1) Abyb} = (2b)Bya` - Mathematics

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

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Solution

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

  Is there an error in this question or solution?
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