Question

sin4A+sin2A/1+cos2A+cos4A=tan2A (prove this)
sin3A-sinA/cos3A+cosA=tanA (prove)

Answer 1

Conditions

sin4A+sin2A/1+cos2A+cos4A=tan2A (prove this)

sin3A-sinA/cos3A+cosA=tanA (prove)

Solution

$$\begin{array}{l} \frac{\sin 4\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \cos 4\alpha} = \tan 2\alpha \\ \frac{\sin 4\alpha + \sin 2\alpha}{1 + \cos 2\alpha + \cos 4\alpha} = \frac{2 \sin 2\alpha \cos 2\alpha + \sin 2\alpha}{\cos^2 2\alpha + \sin^2 2\alpha + \cos 2\alpha + \cos^2 2\alpha - \sin^2 2\alpha} \\ = \frac{\sin 2\alpha (2\cos 2\alpha + 1)}{\cos 2\alpha (2\cos 2\alpha + 1)} = \tan 2\alpha \\ \end{array}$$ $$\begin{array}{l} \frac{\sin 3\alpha - \sin \alpha}{\cos 3\alpha + \cos \alpha} = \tan \alpha \\ \frac{\sin 3\alpha - \sin \alpha}{\cos 3\alpha + \cos \alpha} = \frac{2 \sin \alpha \cos 2\alpha}{2 \cos 2\alpha \cos \alpha} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha \\ \end{array}$$