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The question is:

Prove that: $$ \cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B} $$

I have tried expanding it as $\dfrac{\cos(A+B)}{\sin(A+B)}$ and $\dfrac{1}{\tan(A+B)}$.

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    $\begingroup$ Did you try to use the identities for $\cos(A+B)$ and for $\sin(A+B)$? $\endgroup$ – Dennis Gulko Jul 9 '13 at 7:55
  • $\begingroup$ I tried the $sin(A+B)$ and the $cos(A+B)$ expansions as well as the $tan(A+B)$. The book that I am working from asks me to prove it. $\endgroup$ – zxzxzx Jul 9 '13 at 8:01
  • $\begingroup$ Hint: use the formula for $\tan(A+B)$. $\endgroup$ – Raymond Manzoni Jul 9 '13 at 8:02
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Witout using the identity for $\tan(A+B)$: $$\cot(A+B)=\frac{\cos(A+B)}{\sin(A+B)}=\frac{\cos A\cos B-\sin A\sin B}{\sin A\cos B+\cos A\sin B}=\frac{\sin A\sin B(\cot A\cot B-1)}{\sin A\sin B(\cot B+\cot A)}$$

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  • $\begingroup$ Thank you for the last step there. I wouldn't have thought of factoring as cot, I was only looking for like terms. I suppose the book would be an error then. $\endgroup$ – zxzxzx Jul 9 '13 at 8:38
  • $\begingroup$ A way of motivating Dennis Gulko's last step is to notice that both numerators, the numerator of Dennis Gulko's next to last step and the numerator of what you're trying to get, are binomial differences. The binomial difference that you're trying to obtain involves subtracting $1,$ so this suggests factoring out $\sin A \sin B$ to get a subtraction of $1$ to show up. Once you do this, you get $\sin A \sin B$ times the numerator you want, which means you want to factor the same thing out of the denominator in order to cancel the common $\sin A \sin B$ terms. $\endgroup$ – Dave L. Renfro Jul 10 '13 at 15:17
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Recall that $$ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} $$ and $$ \cot x=\frac{1}{\tan x} $$ It is remains to perform a little algebra.

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  • $\begingroup$ Thanks as well, this method also works out, but with more algebra. $\endgroup$ – zxzxzx Jul 9 '13 at 8:53
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It is probably easier to start from the RHS:

$$\frac{\cot A\cot B-1}{\cot A+\cot B}= \frac{\frac{\cos A}{\sin A}\frac{\cos B}{\sin B} -1}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B} }$$

Now, all the following computations are natural, and pretty authomatic:

$$=\frac{\frac{\cos A \cos B}{\sin A \sin B}-\frac{\sin A\sin B }{\sin A\sin B} }{\frac{\cos A \sin B+ \sin A \cos B}{\sin A \sin B}}=\frac{\cos A \cos B-\sin A\sin B }{\cos A \sin B+ \sin A \cos B}=\frac{\cos(A+B)}{\sin(A+B)}$$

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  • $\begingroup$ How does the 1 become $\frac{sinAsinB}{cosAcosB}$? $\endgroup$ – zxzxzx Jul 10 '13 at 14:34
  • $\begingroup$ @zxzxzx Typo, fixed, ty. Should had been $\frac{\sin A \sin B}{\sin A \sin B}$. $\endgroup$ – N. S. Jul 10 '13 at 14:42
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(cot B / 1-tan b) + (tan B / 1-cot B) = 1 + sec B csc B

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  • $\begingroup$ Please see this post for help with proper formatting of formulas. $\endgroup$ – user147263 Jul 8 '14 at 4:13

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