If $A+B=90^{\circ}$, then $\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}$ is equal to
(a) $\cot ^{2} \mathrm{~A}$
(b) $\cot ^{2} \mathrm{~B}$
(c) $-\tan ^{2} A$
(d) $-\cot ^{2} A$
We have:
$A+B=90^{\circ}$
$\Rightarrow B=90^{\circ}-A$
We have to find the value of the following expression
$\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}$
So
$\frac{\tan A \tan B+\tan A \cot B}{\sin A \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}$
$=\frac{\tan A \tan \left(90^{\circ}-A\right)+\tan A \cot \left(90^{\circ}-A\right)}{\sin A \sec \left(90^{\circ}-A\right)}-\frac{\sin ^{2}\left(90^{\circ}-A\right)}{\cos ^{2} A}$
$=\frac{\tan A \cot A+\tan A \tan A}{\sin A \operatorname{cosec} A}-\frac{\cos ^{2} A}{\cos ^{2} A}$
$=1+\tan ^{2} A-1$
$=\tan ^{2} A$
$=\tan ^{2}\left(90^{\circ}-B\right)$
$=\cot ^{2} B$
Hence the correct option is $(b)$