Any "shortcuts" to proving that $\frac{\sin(x)}2+\sin^2(\frac x2)\tan(\frac x2)\to\tan(\frac x2)$ I was working on simplifying some trig functions, and after a while of playing with them I simplified $$\frac{\sin(x)}{2}+\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right) \rightarrow \tan\left(\frac{x}{2}\right)$$
The way I got that result, however, was with what I think a very "roundabout" way. I first used the half-angle formulaes, then used $x=\pi/2-\beta$, and that simplified to $$\frac{\cos(\beta)}{1+\sin(\beta)}$$ where I again used the coordinate change to get $$\frac{\sin(x)}{1+\cos(x)}\rightarrow\tan\left(\frac{x}{2}\right)$$
I tried using the online trig simplifiers but none succeeded. Of course, after you know the above identity, it's easy to prove by proving that $$\frac{\sin(x)}{2}=\tan\left(\frac{x}{2}\right)-\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)$$
Is there a more direct way to get the identity? I guess what I'm asking is, am I missing any "tricks" or software that I could have on my toolbelt so that next time I don't spend hours trying to simplify trig identities?

Fairly obvious: $$\sin^2\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=(1-\cos^2\left(\frac{x}{2}\right))\tan\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right)=\tan\left(\frac{x}{2}\right)-\frac{\sin x}{2}$$
