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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.

4 Answers 4

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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

$\frac{\cos(A)}{1-\tan(A)} + \frac{\sin(A)}{1-\cot(A)} = \frac{\cos(A)}{1-\frac{\sin(A)}{\cos(A)}} + \frac{\sin(A)}{1-\frac{\cos(A)}{\sin(A)}} = \frac{\cos^2(A)}{\cos(A) - \sin(A)} - \frac{\sin^2(A)}{\cos(A) - \sin(A)} = \cos(A) + \sin(A)$

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    I wasn't aware that $\frac{x}{1 - \frac{y}{x}} = -\frac{x^2}{y - x}$. I thought I knew algebra fairly well! Thank you very much. Case closed!2011-03-07
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To start, I'd suggest rewriting the left side in terms of sine and cosine (the tangent and cotangent) and simplifying the complex fractions.

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    Thanks for the explanation. It's handy to have that in the toolbox to clear an inner fraction.2011-03-08
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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.

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    It *would* be humorous to see the teacher's reaction to such a solution. Of course, you might get a reprimand for cheating!2011-03-10
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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 $\cos$.