Prove the trigonometric identity: ( heta + 2 heta) / (1 + heta + 2 heta) = heta

We will use the following identities:

• Double angle identity for sine: $\sin 2\theta = 2 \sin \theta \cos \theta$
• Double angle identity for cosine: $\cos 2\theta = 2 \cos^2 \theta - 1$
• Tangent identity: $\tan \theta = \frac{\sin \theta}{\cos \theta}$

The numerator is $\sin \theta + \sin 2\theta$.
Substitute $\sin 2\theta = 2 \sin \theta \cos \theta$:

$$\sin \theta + \sin 2\theta = \sin \theta + 2 \sin \theta \cos \theta$$

Factor out $\sin \theta$:

$$= \sin \theta (1 + 2 \cos \theta)$$

The denominator is $1 + \cos \theta + \cos 2\theta$.
Substitute $\cos 2\theta = 2 \cos^2 \theta - 1$:

$$1 + \cos \theta + \cos 2\theta = 1 + \cos \theta + (2 \cos^2 \theta - 1)$$

Combine like terms:

$$= \cos \theta + 2 \cos^2 \theta$$

Factor out $\cos \theta$:

$$= \cos \theta (1 + 2 \cos \theta)$$

Given that $\cos \theta \neq -\frac{1}{2}$, we know that $1 + 2 \cos \theta \neq 0$, so we can cancel out the common term $(1 + 2 \cos \theta)$:

$$= \frac{\sin \theta}{\cos \theta}$$

Using the tangent identity, and given that $\cos \theta \neq 0$:

$$= \tan \theta$$

Thus, the left-hand side simplifies to $\tan \theta$, which is equal to the right-hand side.

$$\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = \tan \theta$$

The identity is proven.