Trig identity proof help I'm trying to prove that
$$ \frac{\cos(A)}{1-\tan(A)} + \frac{\sin(A)}{1-\cot(A)} = \sin(A) + \cos(A)$$
Can someone help me to get started? I've done other proofs but this one has me stumped! Just a start - I don't need the whole proof. Thanks.
 A: Hint: Write $\tan(A)$ as $\frac{\sin(A)}{\cos(A)}$ and $\cot(A)$ as $\frac{\cos(A)}{\sin(A)}$ do some algebra and use the fact that $\frac{x^2-y^2}{x-y} = x + y$.
Note that the equality makes sense only when $A \neq n \pi, n \pi + \frac{\pi}{4}, n \pi + \frac{\pi}{2}$
EDIT
$\frac{\cos(A)}{1-\tan(A)} + \frac{\sin(A)}{1-\cot(A)} = \frac{\cos(A)}{1-\frac{\sin(A)}{\cos(A)}} + \frac{\sin(A)}{1-\frac{\cos(A)}{\sin(A)}} = \frac{\cos^2(A)}{\cos(A) - \sin(A)} - \frac{\sin^2(A)}{\cos(A) - \sin(A)} = \cos(A) + \sin(A)$
A: To start, I'd suggest rewriting the left side in terms of sine and cosine (the tangent and cotangent) and simplifying the complex fractions.
A: You can use the complex variable $q=e^{iA}$ and the following identities
$$\cos A=\frac{e^{iA}+e^{-iA}}{2},\qquad\sin A=\frac{e^{iA}-e^{-iA}}{2i},$$
$$\tan A=\frac{e^{iA}-e^{-iA}}{i\left( e^{iA}+e^{-iA}\right) }\qquad\text{and}\qquad\cot A=\frac{i\left( e^{iA}+e^{-iA}\right) }{e^{iA}-e^{-iA}},$$
to get:
$$\dfrac{\dfrac{q+q^{-1}}{2}}{1-\dfrac{q-q^{-1}}{i\left( q+q^{-1}\right) }}+%
\dfrac{\dfrac{q-q^{-1}}{2i}}{1-\dfrac{i\left( q+q^{-1}\right) }{q-q^{-1}}}=%
\dfrac{q-q^{-1}}{2i}+\dfrac{q+q^{-1}}{2},$$
which you may then verify algebraically that is an identity.
Note: the denominators in the original identity have to be different from zero and  $\tan A$ and $\cot A$ cannot be infinity.

Added: the last formula can be simplified as follows. Firstly 
$$\dfrac{-\left( q+q^{-1}\right) ^{2}}{i\left( q+q^{-1}\right) -q+q^{-1}}-%
\dfrac{\left( q-q^{-1}\right) ^{2}}{-q+q^{-1}+i\left( q+q^{-1}\right) }%
=q-q^{-1}+i\left( q+q^{-1}\right) ,$$
then
$$-\left( q+q^{-1}\right) ^{2}-\left( q-q^{-1}\right) ^{2}$$
$$=\left( q-q^{-1}+i\left( q+q^{-1}\right) \right) \left( i\left(
q+q^{-1}\right) -q+q^{-1}\right) ,$$
and finally 
$$-2q^{2}-\frac{2}{q^{2}}=-2q^{2}-\frac{2}{q^{2}},$$
which is indeed an identity.
A: I would try multiplying the numerator and denominator both by (for the first term) $1+\tan{(A)}$ and for the second $1+\cot{(A)}$.  From there it should just be a little bit of playing with Pythagorean identities ($\sin^2{(A)}+\cos^2{(A)}=1$, $\tan^2{(A)}+1=\sec^2{(A)}$, and $1+\cot^2{(A)}=\csc^2{(A)}$) and writing $\tan{(A)}$ and $\cot{(A)}$ in terms of $\sin$ and $\cos$.
