# How do i prove this trigonometric expression?

How do you prove this? $${(1-2\sin^2A)^2 \over \cos^4A-\sin^4A} = {2\cos^2A - 1}$$

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Express everything in terms of $u = \cos A$, and go from there... – vonbrand May 9 '13 at 18:14
I don't.${}{}{}$ – Michael Greinecker May 9 '13 at 21:12

\begin{align} \frac{(1-2\sin^2(A))^2}{\cos^4(A)-\sin^4(A)} &=\frac{(1-2\sin^2(A))^2}{(1-\sin^2(A))^2-\sin^4(A)}\\ &=\frac{(1-2\sin^2(A))^2}{1-2\sin^2(A)}\\ &=1-2\sin^2(A)\\[9pt] &=1-2(1-\cos^2(A))\\[9pt] &=2\cos^2(A)-1 \end{align}

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Recall the following expressions for $\cos(2A)$: $$\color{magenta}{\cos(2A)} = \cos^2(A) - \sin^2(A) = \color{red}{2\cos^2(A)-1} = \color{blue}{1-2\sin^2(A)} = \color{green}{\cos^4(A) - \sin^4(A)}$$ Hence, we have $$\dfrac{\left(\color{blue}{1-2\sin^2(A)}\right)^2}{\color{green}{\cos^4(A) - \sin^4(A)}} = \dfrac{\color{magenta}{\cos^2(2A)}}{\color{magenta}{\cos(2A)}} = \color{magenta}{\cos(2A)} = \color{red}{2\cos^2(A)-1}$$

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${(1-2\sin^2A)^2 \over \cos^4A-\sin^4A}$

$cos^22A \over \ (cos^2A-\sin^2A)(cos^2A+\sin^2A)$

$cos^22A \over \ (cos^2A-\sin^2A) \ *\ 1$

$cos^22A \over \ (cos2A)$

cos2A

${2\cos^2A - 1}$

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Observe that we need to eliminate $\sin A$

So using $\sin^2A=1-\cos^2A,$

$1-2\sin^2A=1-2(1-\cos^2A)=2\cos^2A-1$

and $\cos^4A-\sin^4A=(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)=1\cdot\{\cos^2A-(1-\cos^2A)\}=2\cos^2A-1$

or $\cos^4A-\sin^4A=\cos^4A-(1-\cos^2A)^2=2\cos^2A-1$

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