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antoniarose · Sep 5, 2011, 01:15 PM

I have tried so many times to figure out how to verify this identity! I need all the help I can get! Please help! I have tried so much! If you can tell me the steps to verify it, that would be SO HELPFUL!

SecB * cscB - 2(cosB * cscB) + cotB = tanB

jcaron2 · Sep 5, 2011, 08:19 PM

Start by rewriting everything in terms of sines and cosines:

$(\sec B)(\csc B) - 2(\cos B)(\csc B) + \cot B = \tan B$

$(\frac{1}{\cos B})(\frac{1}{\sin B}) - 2(\cos B)(\frac{1}{\sin B}) + \frac{\cos B}{\sin B} = \frac{\sin B}{\cos B}$

Now let's get a common denominator for the left side:

$(\frac{1}{\cos B})(\frac{1}{\sin B}) - 2(\cos B)(\frac{1}{\sin B})(\frac{\cos B}{\cos B}) + \frac{\cos B}{\sin B}(\frac{\cos B}{\cos B}) = \frac{\sin B}{\cos B}$

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

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

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

Now remember that $1-\cos^2 B = \sin^2 B$ , and substitute that into the numerator on the left side:

$\frac{\sin^2 B}{(\cos B)(\sin B)} = \frac{\sin B}{\cos B}$

Finally, just cancel out the extra sine terms in the numerator and denominator:

$\frac{\sin B}{\cos B} = \frac{\sin B}{\cos B}$

There you have it! Did that all make sense?