1. ## identity

Prove (Cos 2Θ)/(1+sin2Θ)= (cotΘ -1)/(cotΘ+1)

2. Here.

3. Hello, Godfather!

Prove: .$\displaystyle \frac{\cos2\theta}{1+\sin2\theta} \:= \:\frac{\cot\theta -1}{\cot\theta +1}$

The right side is: .$\displaystyle \frac{\dfrac{\cos\theta}{\sin\theta} - 1}{\dfrac{\cos\theta}{\sin\theta} + 1}$

Multiply by $\displaystyle \frac{\sin\theta}{\sin\theta}\!:\quad \frac{{\color{blue}\sin\theta}\left(\dfrac{\cos\th eta}{\sin\theta} - 1\right)}{{\color{blue}\sin\theta}\left(\dfrac{\co s\theta}{\sin\theta} + 1\right)} \;=\;\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}$

Multiply by $\displaystyle \frac{\cos\theta+\sin\theta}{\cos\theta +\sin\theta}\!:\quad {\color{blue}\frac{\cos\theta + \sin\theta}{\cos\theta + \sin\theta}} \cdot \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\;=$ .$\displaystyle \frac{\cos^2\!\theta -\sin^2\!\theta}{\cos^2\!\theta + 2\sin\theta\cos\theta + \sin^2\!\theta}$

. . . $\displaystyle = \;\frac{\overbrace{\cos^2\!\theta - \sin^2\!\theta}^{\text{This is }\cos2\theta}}{\underbrace{\sin^2\!\theta + \cos^2\!\theta}_{\text{This is 1}} + \underbrace{2\sin\theta\cos\theta}_{\text{This is }\sin2\theta}} \;=\;\frac{\cos2\theta}{1 + \sin2\theta} \quad \hdots \text{ ta-}DAA!$