Verified answer

[tex] \frac{1 - 4 \sin {}^{2} ( \alpha ) \cos {}^{2} ( \alpha ) }{ \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) } = \frac{1 -( 2 \sin( \alpha ) \cos( \alpha )) {}^{2} }{ \cos(2a) } = \\ \frac{1 - \sin {}^{2} (2 \alpha ) }{ \cos( 2\alpha ) } = \frac{ \sin {}^{2} ( 2\alpha ) + \cos {}^{2} (2 \alpha ) - \sin {}^{2} (2 \alpha ) }{ \cos(2 \alpha ) } = \\ \frac{ \cos {}^{2} (2 \alpha ) }{ \cos( 2\alpha ) } = \cos( 2\alpha ) [/tex]