Prove $\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$ $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}=1$$
Prove that L.H.S.$=$R.H.S.

This type of questions always creates problem when in right hand side some trigonometry function is given then it is bit easy to think how to proceed further.
Can some one help me not only to solve the problem but also how to tackle this type of other problem (when R.H.S. is $1$)?

HINT
Simplify as follows : substitute $\sec \theta = \frac{1}{\cos \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$. By this, we will get first term of LHS is 
$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$. Similar if we substitute \second term of LHS, we will get LHS =
$\frac{\sin \theta \ \cos \theta }{1 + \sin \theta - \cos \theta }$ + $\frac{\sin \theta \ \cos \theta }{1 + \cos \theta - \sin \theta }$
$= \sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 + \cos \theta - \sin \theta } ]$
$=\sin \theta \ \cos\theta [\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$
Now simplifying $[\frac {1}{1 + \sin \theta - \cos \theta} + \frac {1}{1 - (\sin \theta - \cos \theta) } ]$ we get $\frac {1}{\cos \theta \sin \theta}$. But it is with product of $\sin \theta \cos \theta$. Hence the answer is $1$, Which is LHS. Hence proved.  
And we should tackle these problems always by substituting in $\sin \theta$ and $\cos \theta$ when LHS is a number.
