This looks like homework, possibly written to encourage interpretation of phase shifts in trigonometric expressions.
tan^2(7pi/2+a)[1-(sin(a-pi/2))^2]+(cos(3pi/2+a))^2 TeXses out like so:
$$\tan ^2\left(\frac{7 \pi }{2}+a\right) \left(1-\sin ^2\left(a-\frac{\pi }{2}\right)\right)+\cos ^2\left(\frac{3 \pi
}{2}+a\right)$$
Simplifying via:
$\tan(x+\pi / 2)=\tan(x)^{-1}$,
$\cos^2(x)+sin^2(x)=1$,
$\sin(x+\pi/2)=\cos(x)$.
$\tan ^2\left(\frac{7 \pi }{2}+a\right)=\tan^{-2}(a)$
$1-\sin ^2\left(a-\frac{\pi }{2}\right)=1-\cos ^2(a)=\sin^2(a)$
$\cos ^2\left(\frac{3 \pi }{2}+a\right)=\sin ^2(a)$
$$\tan^{-2}(a)\sin^2(a)+\sin ^2(a)=\cos^{2}(a)+\sin ^2(a)=1$$