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I have a major problem with verifying a Trigonometric identity. (My teacher couldn't really get it, so I would like to find it just in case it appears on a test)

The problem goes like this:

$$\sin(\theta) + \cos(\theta) = \frac{\sin(\theta)}{1 - \cot(\theta)} + \frac{\cos(\theta)}{1 - \tan(\theta)}$$

Where I must verify that this is true. I have tried numerous operations, mostly getting to $\sin + \cos$ along with a crazy fraction in the numerator and another fraction in the numerator. As I have said, my teacher showed my something, but it really did not work after I continued simplifying.

If there is anyone who would help guide me in this, it would be much appreciated.

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It probably is solved by multiplying the right side by a trigonometric function over the same function (1) in order to create a Pythagorean identity, but I have found no solution by doing this. – nmagerko Apr 4 '12 at 21:39
$1-\tan\theta=1-\cot\theta$ is $0$ when $\theta=\pi/4$ where as the LHS is well-defined for all values of $\theta$.... – user21436 Apr 4 '12 at 21:41
I cannot solve in order to prove. I have to say something like $\sin + \cos$ = $\sin + \cos$ . Otherwise, I would agree with you. – nmagerko Apr 4 '12 at 21:43
@Kanna: If we take the RHS in its totality we see $\pi/4$ is a removable singularity, so we ignore it. – anon Apr 4 '12 at 21:46
When in doubt fall back on Euler. It may blow up into some very complicated algebra but it will cut through all the identities like a knife. – Ben Jackson Apr 5 '12 at 0:10
up vote 9 down vote accepted

$$\eqalign{ {\sin\theta\over 1-\cot\theta} +{\cos\theta\over 1-\tan\theta} &={\sin\theta\over 1-{\cos\theta\over \sin\theta}} +{\cos\theta\over 1-{\sin\theta\over \cos\theta}}\cr &={\sin^2\theta\over \sin\theta-{\cos\theta }} +{\cos^2\theta\over\cos\theta-{\sin\theta }} \phantom{{a\over b}\over{c\over d}}\cr &={\sin^2\theta-\cos^2\theta\over \sin\theta-{\cos\theta }}\phantom{{a\over b}\over{c\over d}} \cr &={(\sin\theta +\cos)(\sin\theta -\cos\theta)\over \sin\theta-{\cos\theta }}\phantom{{a\over b}\over{c\over d}}\cr &=\sin\theta+\cos\theta . } $$

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I apologize for my ignorance, but how did you go from your second step to the third? When I try for a common denominator, I have something different. – nmagerko Apr 4 '12 at 21:50
@nmagerko: Change signs in the second term: $${\cos^2\theta\over\cos\theta-{\sin\theta }} = {\color{Red}- \cos^2\theta\over {\sin\theta -\cos\theta}}.$$ – anon Apr 4 '12 at 21:52
... Oops. Alright, thank you all very much. – nmagerko Apr 4 '12 at 21:53

It bears a vague resemblance to the Pythagorean identity (a sine added to a cosine), and the definitions of $\cot$ and $\tan$ will allow us to realize a route forward by doubling up on them up top:

$$\frac{\sin}{1-\cot} + \frac{\cos}{1-\tan} = \frac{\sin^2}{\sin-\cos}+\frac{\cos^2}{\cos-\sin}=\frac{\sin^2-\cos^2}{\sin-\cos}.$$

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Thank you very much for your answer, however, I must accept the first answer given. I trust you'll understand. – nmagerko Apr 4 '12 at 21:54

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