Ans: C
Since $A+B=90^\circ$, then $B = 90^\circ – A$.
Since $A+B=90^\circ$, then $B = 90^\circ – A$.
$\begin{array}{cl}
& \cos^2 A + \sin^2 B \\
= & \cos^2 A + [\sin (90^\circ – A)]^2\\
= & \cos^2 A + \cos^2 A \\
= & 2 \cos^2 A
\end{array}$
