not able to prove a trig identity For the identity,
$$\dfrac{\cos A}{1 - \tan A} + \dfrac{\sin A}{1 - \cot A}=\sin A + \cos A$$
What I have been able to perform,
$$
\dfrac{\cos A}{\dfrac{\cos A}{\cos A}-\dfrac{\sin A}{\cos A}} + \dfrac{\sin A}{\dfrac{\sin A}{\sin A} - \dfrac{\cos A}{\sin A}}$$
$$
\dfrac{\cos^2 A}{\cos A - \sin A} + \dfrac{\sin^2 A}{\sin A - \cos A}
$$
I am unable to prove this identity
 A: Hint:
$${c^2\over c-s}+{s^2\over s-c}={s^2-c^2\over s-c}$$
Now factor the numerator.  (In other words, what you did was correct, you just have to keep going.)
A: Okay, so we want to verify that $$\frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} = \sin{A} + \cos{A}.$$
Staring with the left hand side, we have (based on what you've already done):
\begin{split} \frac{\cos{A}}{1 - \tan{A}} + \frac{\sin{A}}{1 - \cot{A}} &= \frac{\cos{A}}{\left(\frac{\cos{A} - \sin{A}}{\cos{A}}\right)} + \frac{\sin{A}}{\left(\frac{\sin{A} - \cos{A}}{\sin{A}}\right)} \\ &= \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} + \frac{\sin^{2}{A}}{\sin{A}-\cos{A}}  \end{split}
where we got the last step by inverting and multiplying the fractions in the denominators. Now, it should be clear to you that $\sin{A}- \cos{A}= -(\cos{A} - \sin{A})$ so we can make the denominators of the two fractions the same by pulling out a minus sign from the second denominator, to get:
\begin{split} \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} + \frac{\sin^{2}{A}}{\sin{A}-\cos{A}} &= \frac{\cos^{2}{A}}{\cos{A}-\sin{A}} - \frac{\sin^{2}{A}}{\cos{A}-\sin{A}} \\ &= \frac{\cos^{2}{A} - \sin^{2}(A)}{\cos{A}-\sin{A}}  \end{split}
and recognizing the numerator of that last fraction is the difference of squares! So it factors into $(\cos{A} - \sin{A})(\cos{A} + \sin{A})$, giving us:
\begin{split}\frac{\cos^{2}{A} - \sin^{2}(A)}{\cos{A}-\sin{A}} &= \frac{(\cos{A} - \sin{A})(\cos{A} + \sin{A})}{\cos{A}-\sin{A}} \\ &= \cos{A} + \sin{A} \end{split}
where we cancelled the common factor of $(\cos{A} - \sin{A})$ from the numerator and denominator in the last step.
