<s>[INST] 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 $1$)?
 [/INST] $$\frac{\sin A}{\sec A+\tan A-1}+ \frac{\cos A}{\csc A+\cot A-1}= \frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-\frac{\cos A}{\cos A}}+ \frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}} =$$ $$= \frac{\sin A \cos A}{1 + \sin A - \cos A} + \frac{\sin A \cos A}{1 + \cos A - \sin A} =$$ $$= \sin A\cos A\frac{1 + \sin A - \cos A + 1 + \cos A - \sin A}{(1+\sin A - \cos A)(1 + \cos A - \sin A)} = $$ $$=\frac{2\sin A \cos A}{1 + \cos A - \sin A + \sin A + \sin A\cos A -\sin^2 A -\cos A -\cos^2 A +\sin A\cos A} = $$ $$=\frac{2\sin A \cos A}{2\sin A\cos A} = 1$$
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