Show that: $ (\csc\theta - \sin\theta)(\sec\theta - \cos\theta) \equiv \frac{1}{\tan\theta + \cot\theta}$ I am having hard time solving this question, I start with the L.H.S and my answer always boils down to 
$\sin\theta\cos\theta$
And I don't know what to do after that.

$$ (\csc\theta - \sin\theta)(\sec\theta - \cos\theta) =\frac{(1-\sin^2\theta)(1-\cos^2\theta)}{\sin\theta \cos\theta}= \frac{(\cos^2\theta)(\sin^2\theta)}{\sin\theta \cos\theta}=(\cos\theta)(\sin\theta)=\frac{(\cos\theta)(\sin\theta)}{\sin^2\theta+ \cos^2\theta}=\frac{1}{\tan\theta + \cot\theta}$$
The last inequality comes from dividing numerator  and denominator by the same number $(\cos\theta)(\sin\theta)$.
