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(cot^2B-cos^2B)/csc^2B-1=cos^2B

 May 26, 2014

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 #1
avatar+33654 
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I think this should be written as (cot2B - cos2B)/(csc2B - 1) 

\frac{\cot^2B-\cos^2B}{\csc^2B-1}=\frac{\frac{\cos^2B}{\sin^2B}-\cos^2B}{\frac{1}{sin^2B}-1}=\frac{\cos^2B-\sin^2B\cos^2B}{1-\sin^2B}=\frac{\cos^2B(1-\sin^2B)}{1-\sin^2B}=\cos^2B$$

 May 26, 2014
 #1
avatar+33654 
+5
Best Answer

I think this should be written as (cot2B - cos2B)/(csc2B - 1) 

\frac{\cot^2B-\cos^2B}{\csc^2B-1}=\frac{\frac{\cos^2B}{\sin^2B}-\cos^2B}{\frac{1}{sin^2B}-1}=\frac{\cos^2B-\sin^2B\cos^2B}{1-\sin^2B}=\frac{\cos^2B(1-\sin^2B)}{1-\sin^2B}=\cos^2B$$

Alan May 26, 2014

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