cos2A/(1+sin2A) = (1-tanA)/(1+tanA) | Prove | Trigonometry

Question: Prove that: $\dfrac{cos2A}{1+ sin2A} = \dfrac{1- tanA}{1+ tanA}$

Solution:

Taking LHS,

$= \dfrac{cos2A}{1+ sin2A}$

Expanding cos2A and sin2A and identity of 1

$= \dfrac{cos^2A - sin^2A}{sin^2A + cos^2A + sin2A}$

Performing simple mathematical operations

$= \dfrac{(cosA + sinA)(cosA - sinA)}{(cosA + sinA)^2}$

$= \dfrac{cosA - sinA}{cosA + sinA}$

Dividing each term by cosA

$= \dfrac{\frac{cosA}{cosA} - \frac{sinA}{cosA} }{\frac{cosA}{cosA} + \frac{sinA}{cosA}}$

We know, [$\frac{sinA}{cosA} = tanA$]

$= \dfrac{1 - tanA}{1 + tanA}$

= RHS

#proved