Please help me prove this trig identity (please include justification)
(sinA^4 - cosA^4) secA^2 =secA^2-2
appreciated
Are you sure that that's not supposed to be $\displaystyle \left(\sin^4A-\cos^4A\right)\sec^2A = \sec^2A-2\mathrm?$
If that is what you are trying to prove, begin by factoring the $\displaystyle \sin^4A-\cos^4A$ as a difference of squares. Then use the Pythagorean identity and the rest should be obvious.
Thank you.
This is your error. We cannot distribute powers over addition or subtraction and, likewise, we cannot "factor out" a power like that. $\displaystyle a^2+b^2$ is not necessarily equal to $\displaystyle (a+b)^2$. Instead, convert the secants to cosines:
$\displaystyle \left(\sin^4A - \cos^4A\right)\sec^2A$
$\displaystyle =\left(\sin^2A + \cos^2A\right)\left(\sin^2A - \cos^2A\right)\sec^2A$
$\displaystyle =\left(\sin^2A - \cos^2A\right)\sec^2A$
$\displaystyle =\frac{\sin^2A - \cos^2A}{\cos^2A}$
$\displaystyle =\frac{\sin^2A}{\cos^2A} - \frac{\cos^2A}{\cos^2A}$
Can you continue from here?