अगर A + B = 90° तो, $\sqrt{\frac{\tan A \cdot \tan B+\tan A \cdot \cot B}{\sin A \cdot \operatorname{secB}}-\frac{\ पाप ^{2} बी} {\cos ^{2} ए}}=?$

$A+B=90^{\circ} \Rightarrow B \Rightarrow 90^{\circ}-A$

$\therefore \sqrt{\frac{\tan A \cdot \tan B+\tan A \cdot \cot B}{\sin A \cdot \sec B}-\frac{\sin ^{2} B}{\cos ^{2} A}}$

$=\sqrt{\frac{\tan A \cdot \tan \left(90^{\circ}-\mathrm{A}\right)+}{\tan A \cdot \cot \left(90^{\circ}-\mathrm{A}\right)}-\frac{\sin ^{2}\left(90^{\circ}-\mathrm{A}\right)}{\cos ^{2} \mathrm{~A}}}$

$=\sqrt{\frac{\tan A \cdot \cot A+\tan A \cdot \cot A}{\sin A \cdot \operatorname{cosec} A}-\frac{\cos ^{2} A}{\cos ^{2} A}}$

$=\sqrt{1+\tan ^{2} A-1}=\sqrt{\tan ^{2} A}=\tan A .$