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.


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 $A \neq n \pi, n \pi + \frac{\pi}{4}, n \pi + \frac{\pi}{2}$


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

  • $\begingroup$ Thanks for your reply. I completed the proof but I suspect it was unnecessarily complex. Started with $\frac{\cos(A)}{1 - \frac{\sin(A)}{\cos(A)}} + \frac{\sin(A)}{1 - \frac{\cos(A)}{\sin(A)}}=\sin(A) + \cos(A)$. Then multipled both sides by $1-\frac{\sin(A)}{\cos(B)}$, then multipled both sides by $1-\frac{\cos(A)}{sin(A)}$. It got very messy but managed to get there. I didn't get to make use of $\frac{x^2 -y^2}{x - y} = x + y$. I'm sure there must be an easier way but after hours of looking at this, I can't see it :( $\endgroup$ – PeteUK Mar 7 '11 at 12:25
  • $\begingroup$ 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! $\endgroup$ – PeteUK Mar 7 '11 at 17:35

To start, I'd suggest rewriting the left side in terms of sine and cosine (the tangent and cotangent) and simplifying the complex fractions.

  • $\begingroup$ Thanks Isaac. Expressing in sin and cos is what my text recommends too. My algrbraic manipulation confidence has taken a dent (see my comment to Sivaram's answer). $\endgroup$ – PeteUK Mar 7 '11 at 12:30
  • $\begingroup$ @PUK: Looking at your comment there, it seems like the simplification of the complex fractions is the issue. Regarding $\frac{x}{1 - \frac{y}{x}} = -\frac{x^2}{y - x}$, multiplying the numerator and denominator of the overall fraction by the same quantity is multiplying the fraction by 1, which does not change the fraction, and multiplying the numerator and denominator by the denominator of the inner fraction clears the inner fraction: $$\frac{x}{1-\frac{y}{x}}=\frac{x}{1-\frac{y}{x}}\cdot\frac{x}{x} =\frac{x^2}{x-\frac{y}{x}\cdot x}=\frac{x^2}{x-y}$$ $$=\frac{x^2}{-(y-x)}=-\frac{x^2}{y-x}$$ $\endgroup$ – Isaac Mar 7 '11 at 18:22
  • $\begingroup$ Thanks for the explanation. It's handy to have that in the toolbox to clear an inner fraction. $\endgroup$ – PeteUK Mar 8 '11 at 16:22

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

$$\dfrac{-\left( q+q^{-1}\right) ^{2}}{i\left( q+q^{-1}\right) -q+q^{-1}}-% \dfrac{\left( q-q^{-1}\right) ^{2}}{-q+q^{-1}+i\left( q+q^{-1}\right) }% =q-q^{-1}+i\left( q+q^{-1}\right) ,$$


$$-\left( q+q^{-1}\right) ^{2}-\left( q-q^{-1}\right) ^{2}$$

$$=\left( q-q^{-1}+i\left( q+q^{-1}\right) \right) \left( i\left( q+q^{-1}\right) -q+q^{-1}\right) ,$$

and finally


which is indeed an identity.

  • $\begingroup$ Thank you. I hope to have the skills to look back on this answer and understand it. At the moment my poor brain's struggling with (what I think is) simpler stuff. $\endgroup$ – PeteUK Mar 7 '11 at 12:33
  • $\begingroup$ All the above relations are derived from the Euler's formula ( en.wikipedia.org/wiki/Euler's_formula ) $e^{ix}=\cos x+i\sin x$ which you will probably learn later. $\endgroup$ – Américo Tavares Mar 7 '11 at 14:18
  • $\begingroup$ It would be humorous to see the teacher's reaction to such a solution. Of course, you might get a reprimand for cheating! $\endgroup$ – The Chaz 2.0 Mar 10 '11 at 18:33

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 $\cos$.


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