Question

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

Answer 1

$RTP: \frac{1 + cos(2A)}{cos(2A)} = \frac{tan(2A)}{tan(A)}$

$LHS = \frac{1 + cos(2A)}{cos(2A)}$
$= \frac{1 + \frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}$
$= \frac{\frac{1 + tan^{2}(A) + 1 - tan^{2}(A)}{1 + tan^{2}(A)}}{\frac{1 - tan^{2}(A)}{1 + tan^{2}(A)}}$
$= \frac{2}{1 - tan^{2}(A)}$
$= \frac{2}{1 - tan^{2}(A)} \cdot \frac{tan(A)}{tan(A)}$
$= \frac{\frac{2tan(A)}{1 - tan^{2}(A)}}{tan(A)}$
$= \frac{tan(2A)}{tan(A)}$
$= RHS$, as required.