Question

1+cos2A/cos2A = tan2A/tanA prove LHS=RHS

Answer 1

`RTP: (1 + cos(2A))/(cos(2A)) = (tan(2A))/(tan(A))`


`LHS = (1 + cos(2A))/(cos(2A))`

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

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

`= (2)/(1 - tan^(2)(A))`

`= (2)/(1 - tan^(2)(A)) \cdot (tan(A))/(tan(A))`

`= ((2tan(A))/(1 - tan^(2)(A)))/(tan(A))`

`= (tan(2A))/(tan(A))`

`= RHS`, as required.