Prove that $\tan(x)\tan(x+r) = \frac{\tan(x+r) - \tan(x)}{\tan r} -1$ $\tan(x)\cdot \tan(x+r) = \dfrac{\tan(x+r) - \tan(x)}{\tan r} -1$
I tried L.H.S = $\dfrac{\sin(x)\cdot \sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x+r- x)}{\sin(r)}\dfrac{\sin(x)\cdot\sin(x+r)}{\cos(x)\cdot\cos(x+r)}$
= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\dfrac{\sin(x+r)\cos(x) - \cos(x+r)\cos(x)}{\cos(x)\cdot \cos(x+r)}$
= $\dfrac{\sin(x)\cdot\sin(x+r)}{\sin(r)}\left(\tan(x+r) -\tan(x)\right)$
I get stuck here; what else is there to do?
Edit: using Tavish's hint, we can get it directly.
$$\tan(x+r -x) = \dfrac{\tan(x+r) - \tan(x)}{1 + \tan(x+r)\tan(x)}$$
$$1 + \tan(x+r)\tan(x)= \dfrac{\tan(x+r) - \tan(x)}{\tan(r)}$$
$$\tan(x+r)\tan(x) = \dfrac{\tan(x+r) - \tan(x)}{\tan(r)} -1$$

Hint:
Use $$\tan(a-b)=\frac{\tan a -\tan b}{1+\tan a \tan b} $$ with $a=x+r, b=x$.
