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Can anyone help me verify the following trig identity:

$$(1-\cos A)(1+\sec A)(\cot A)= \sin A.$$

My work so far is

$$(1-\cos A)\left(1+\frac{1}{\cos A}\right)\frac{\cos A}{\sin A}=\sin A$$

But I am stuck around this part.

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You should have $\sec A = 1/\cos A$. If you make this correction, you should be able to get your equation to $1 - \cos^2 A = \sin^2 A$, which is a known identity ($\cos^2 A + \sin^2 A = 1$). – Henry T. Horton Jul 14 '12 at 22:28
yes I meant sec=1/cos I just wrote my work part wrong. – Fernando Martinez Jul 14 '12 at 22:33

Express everything on the left in terms of sines and cosines. We get $$(1-\cos A) \left(1+\frac{1}{\cos A}\right)\frac{\cos A}{\sin A}.$$

First multiply $1+\frac{1}{\cos A}$ by the $\cos A$ in $\frac{\cos A}{\sin A}$. We get $\cos A+1$.Our expression is now equal to $$(1-\cos A)(\cos A+1)\frac{1}{\sin A}.$$ But $(1-\cos A)(\cos A+1)=(1-\cos A)(1+\cos A)=1-\cos ^2 A=\sin^2 A$. Now multiply by the $\frac{1}{\sin A}$. We get $\sin A$.

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thanks for the help. – Fernando Martinez Jul 14 '12 at 22:35
@Rakishi: You are welcome. As you can see, it is a minor slip that prevented you from pushing the argument through. – André Nicolas Jul 14 '12 at 22:39

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