Is this identity provable? I was given this question by a friend but it seems the question is not correctly phrased in the book that he got it.
$ \frac{\cos A}{1-\tan A} +\frac{\sin A}{1-\cot A}=\sin A +\cos A$
Is the above identity provable? 
 A: Let's take left hand side 
$$\color{blue}{\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}}$$ $$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$$
$$=\frac{\cos^2 A}{\cos A-\sin A}+\frac{\sin^2 A}{\sin A-\cos A}$$
$$=\frac{\cos^2 A}{\cos A-\sin A}-\frac{\sin^2 A}{\cos A-\sin A}$$
$$=\frac{\cos^2 A-\sin^2 A}{\cos A-\sin A}$$
$$=\frac{(\cos A+\sin A)(\cos A-\sin A)}{\cos A-\sin A}=\cos A+\sin A$$
$$\color{blue}{=\sin A+\cos A}$$
Hence, the trigonometric identity is correct & provable. 
A: It is correct.
Use these identities: $\tan A=\frac{\sin A}{\cos A},\cot A=\frac{\cos A}{\sin A}$; then multiply and divide it with $\sin A+\cos  A$, use $a^2-b^2=(a+b)(a-b)$
A: $$\frac{\cos^2 A}{\left(\cos A-\sin A\right)}+\frac{\sin^2A}{\left(\sin A-\cos A\right)}$$ is true by factorizing $\frac{1}{(\cos A-\sin A)}$
A: it is an identity. here is one way to prove it:
$$\frac{\cos t }{1- \tan t} + \frac{\sin t}{1 - \cot t} =\frac{\cos^2 t}{\cos t - \sin t} + \frac{\sin^2 t }{\sin t - \cos t} = \frac{\cos^2 t - \sin^2 t}{\cos t - \sin t} = \cos t + \sin t.  $$
