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(cotalpha cotbeta -1)/(cotalpha +cotbeta )

 Dec 6, 2018
 #1
avatar+203 
+2

Do you want it simplified? It already is. What are you looking for in this expression?

 Dec 6, 2018
 #2
avatar+101741 
+1

\(\frac{(cot\alpha cot\beta -1)}{(cot\alpha +cot\beta )}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{sin\alpha sin\beta}\div \left(\frac{cos\alpha}{sin\alpha}+\frac{cos\beta}{sin\beta}\right)\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{sin\alpha sin\beta}\div \frac{2cos\alpha sin\beta}{sin\alpha sin\beta}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{2cos\alpha sin\beta}\\ =\frac{cos\alpha cos\beta-sin\alpha sin\beta}{2cos\alpha sin\beta}\\ =\frac{cos(\alpha+\beta)}{sin(\alpha+\beta)}\\ =cot(\alpha+\beta)\)

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 Dec 7, 2018

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