prove the following identities: \( \begin{array}{ll}\text { (1) } \cos x \sin x+\sin ^{2} x \tan x=\tan x & \text { (2) } \frac{\tan A \cdot \sin 2 A}{1-\cos A}=2+2 \cos A \\ \text { (3) } \frac{\cos \alpha+\sin \alpha}{\cos 2 \alpha}=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos \alpha-\sin \alpha} & \text { (4) } \frac{\sin 2 x-2 \sin x}{\cos 2 x-1}=\frac{1}{\sin x}-\frac{1}{\tan x} \\ \text { (5) } \frac{\sin \theta-\sin 2 \theta}{\cos \theta-\cos 2 \theta-1}=\tan \theta & \text { (6) } \frac{\cos 2 \alpha+\cos \alpha}{\sin 2 \alpha-\sin \alpha}=\frac{\cos \alpha+1}{\sin \alpha}\end{array} \)

Sure! Let's tackle two of the identities. 1. To prove \( \cos x \sin x + \sin^2 x \tan x = \tan x \): Start with the left side: \[ \cos x \sin x + \sin^2 x \tan x = \cos x \sin x + \sin^2 x \frac{\sin x}{\cos x} = \cos x \sin x + \frac{\sin^3 x}{\cos x} \] Combine the terms over a common denominator, which is \( \cos x \): \[ = \frac{\cos^2 x \sin x + \sin^3 x}{\cos x} = \frac{\sin x (\cos^2 x + \sin^2 x)}{\cos x} \] Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \): \[ = \frac{\sin x}{\cos x} = \tan x \] Hence, \( \cos x \sin x + \sin^2 x \tan x = \tan x \) is proven. 2. For the identity \( \frac{\tan A \cdot \sin 2A}{1 - \cos A} = 2 + 2 \cos A \): Start with the left side: \[ \frac{\tan A \cdot \sin 2A}{1 - \cos A} = \frac{\frac{\sin A}{\cos A} \cdot 2 \sin A \cos A}{1 - \cos A} = \frac{2 \sin^2 A}{\cos A (1 - \cos A)} \] We can express \( 1 - \cos A \) in terms of \( \sin^2 A \): \[ 1 - \cos A = \frac{\sin^2 A}{1 + \cos A} \] Therefore: \[ \frac{2 \sin^2 A}{\cos A \cdot \frac{\sin^2 A}{1 + \cos A}} = \frac{2(1 + \cos A)}{\cos A} \] Simplifying gives: \[ = 2 + 2 \cos A \] Thus, the identity is proven.