Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$,using the identity $\operatorname{cosec}^{2} A=1+\cot ^{2} A$.

(N/A) To prove: $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A$
$L.H.S. = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$
Divide the numerator and the denominator by $\sin A$:
$= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}}$
$= \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A}$
Using the identity $1 = \operatorname{cosec}^{2} A - \cot^{2} A$ in the numerator:
$= \frac{\cot A + \operatorname{cosec} A - (\operatorname{cosec}^{2} A - \cot^{2} A)}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) - (\operatorname{cosec} A - \cot A)(\operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Factor out $(\cot A + \operatorname{cosec} A)$:
$= \frac{(\cot A + \operatorname{cosec} A) [1 - (\operatorname{cosec} A - \cot A)]}{\cot A - \operatorname{cosec} A + 1}$
$= \frac{(\cot A + \operatorname{cosec} A) (1 - \operatorname{cosec} A + \cot A)}{\cot A - \operatorname{cosec} A + 1}$
Since $(1 - \operatorname{cosec} A + \cot A) = (\cot A - \operatorname{cosec} A + 1)$,they cancel out:
$= \cot A + \operatorname{cosec} A = R.H.S.$