proving that $\sin\beta \cos(\beta+\theta)=-\sin\theta$ implies $\tan\theta=-\tan\beta$ 
Show that $\sin\beta \cos(\beta+\theta)=-\sin\theta$ implies $\tan\theta=-\tan\beta$

I expand the cosinus:
$$\cos(\beta+\theta)=\left(1-\frac{\theta^2}{2}\right)\left(1-\frac{\beta^2}{2}\right)-\beta\theta$$
but I can't get any tan.
 A: Separating out $\sin\theta,\cos\theta$ after using $$\cos(A+B)$$ formula
$$\sin\theta(\sin^2\beta-1)=\sin\beta\cos\beta\cos\theta$$
$$\iff\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sin\beta\cos\beta}{\sin^2\beta-1}=?$$
A: 
Indeed, 
since $\cos(\beta + \theta) = \cos\beta\cos\theta - \sin\beta\sin\theta$
your equation becomes
 $$
\sin\beta( \cos\beta\cos\theta - \sin\beta\sin\theta) =\sin\theta $$
Dividing by $\cos\theta$ $$
\sin\beta( \cos\beta - \sin\beta\tan\theta) =-\tan\theta \implies 
\sin\beta \cos\beta - \sin^2\beta\tan\theta =-\tan\theta $$
Dividing by $\cos^2\beta = \frac{1}{1+\tan^2\beta}$
$$
\tan\beta - \tan^2\beta\tan\theta =-\tan\theta(1+\tan^2\beta) \implies
\tan\beta =-\tan\theta $$

A: Note that
$$\cos(\beta+\theta)=\cos\beta \cos\theta-\sin\beta \sin\theta $$
thus step by step
$$\sin\beta \cos(\beta+\theta)=\sin\beta \cos\beta \cos\theta-\sin^2\beta \sin\theta=-\sin \theta$$
$$\sin\beta \cos\beta \cos\theta-\sin^2\beta \sin\theta=-\sin \theta$$
$$\sin\beta \cos\beta \cos\theta=-\sin \theta+\sin^2\beta \sin\theta$$
$$\sin\beta \cos\beta \cos\theta=-\sin \theta(1-\sin^2\beta)$$
$$\sin\beta \cos\beta \cos\theta=-\sin \theta \cos^2\beta$$
$$\sin\beta \cos\theta=-\sin \theta \cos\beta$$
$$\frac{\sin\beta} {\cos\beta}=-\frac{\sin \theta} {\cos\theta}\iff\tan\theta=-\tan \beta$$
