Explanation

(c)
Given, $a\sec{\alpha = d + c\tan\alpha}$ …(i)
and $b\sec{\alpha = c - d\tan{\alpha\ \ \ \ \ \ldots(ii)}}$
On squaring and adding Eqs. (i) and (ii), we get
$\left( a^{2} + b^{2} \right)\sec^{2}{\alpha = d^{2} + c^{2}\tan^{2}{\alpha + 2dc\tan{\alpha + c^{2} + d^{2}\tan^{2}{\alpha - 2dc\tan\alpha}}}}$
$\Rightarrow \left( a^{2} + b^{2} \right)\sec^{2}{\alpha = c^{2}\left( \tan^{2}{\alpha + 1} \right) + d^{2}\left( 1 + \tan^{2}\alpha \right)}$
$= \left( c^{2} + d^{2} \right)\sec^{2}\alpha$
$\therefore a^{2} + b^{2} = c^{2} + d^{2}$