Reciprocal Identities Help. $\displaystyle\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A}$ = $\displaystyle\frac{1}{\sec(A)+1}$
Can someone help me with this one? I can't seem to get it right, I get lost, right side has me stumped. Sorry; there's the right side.

Use the identity $\sin^2 x = 1 - \cos^2 x$. Then, factor the left hand side:
$$\frac{\sin^2A + 2\cos A - 1}{2 + \cos A - \cos^2A} = \frac{1 - \cos^2 A+ 2 \cos A - 1}{(2 - \cos A)(1 + \cos A)} = \frac{\cos A(2 - \cos A)}{(2-\cos A)(1 + \cos A)}$$
Now cancel the common factor, to get $$\frac{\cos A}{1 + \cos A}$$
Now, if we divide the numerator and denominator by $\cos A$, the result is $$\frac 1{\frac 1{\cos A} + 1} = \frac 1{\sec A + 1}$$
