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jon lilok · Nov 13, 2012, 08:53 PM

sin theta sec theta/sin squared theta- tan squared theta= cos theta/sin cube theta

ebaines · Nov 14, 2012, 10:13 AM

Please use parentheses, because it's impossible to understand the identity you are trying to prove. I think you may mean this:

$<br /> \frac {\sin ( \theta) sec(\theta)}{\sin^2(\theta) - \tan^2 (\theta)} = \frac {\cos(\theta)}{\sin^3(\theta)}<br />$

I suggest converting the secant and tangent functions to their sine and cosine equivalents, and simplify. You will also have to use the identity sin^2x + cos^2x = 1.