Anyone knows how to prove this equation (using trigonometric identities?) Prove: $$\frac {\sin \beta \cos \beta \pm \sin \alpha \cos \alpha}{\cos^2 \beta - \sin^2 \alpha}=\tan(\beta \pm \alpha)$$
For example: $\beta = 3^\circ $ And $\alpha =23^\circ $, both sides of this equation are $0.4877325$ for the $+$ case. 
I also substituted other different values into this equation and they are all equal, but I do not know how to prove it. It's from a book, just said using trigonometric identities, but I googled a lot, still cannot prove it. Anyone can help me to prove it? Thanks!
 A: If you are willing to start from the identity for the tangent of a sum
$$
\tan(\theta \pm \gamma) = \frac{\tan\theta \pm \tan \gamma}{1\mp \tan\theta \tan \gamma}
$$
then write each $\tan x$ as $\frac{\sin x}{\cos x}$ and then multiply the numerator and denominator by $\cos \theta \cos \gamma$ to get the identity you are trying to prove.
Or you can use the sum of angles formulas for sine and cosine
$$\sin (\theta \pm \gamma) = \sin \theta \cos \gamma  \pm \cos \theta \sin \gamma \\
\cos (\theta \pm \gamma) = \cos\theta\cos\gamma \mp \sin\theta\sin\gamma
$$
and combine them to form $\tan(\theta\pm\gamma)$ getting the same result.
A: It can done using five standard trig identities: $$(1)\qquad\sin2\theta=2\sin\theta\cos\theta,$$ $$(2\text{a,b})\quad\cos2\theta=2\cos^2\theta-1=1-2\sin^2\theta,$$ $$(3)\quad\sin2\theta+\sin2\phi=2\sin(\theta+\phi)\cos(\theta-\phi),$$ $$(4)\quad\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi).$$ Then $$\frac {\sin \beta \cos \beta + \sin \alpha \cos \alpha}{\cos^2 \beta - \sin^2 \alpha}$$ $$=\frac{\sin2\beta+\sin2\alpha}{\cos2\beta+\cos2\alpha}\quad(\text{by identities 1 and 2a,b).}$$ Using identities 3 and 4, and cancelling $\cos(\alpha-\beta)$, now gives the required result for the plus sign. The result for the minus sign is obtained by replacing $\alpha$ by $-\alpha$.
