If $$A + B = 90^\circ$$, then find $$\frac{\sqrt{\tan A \tan B + \tan A \cot B}}{\sin A \sec B}$$.

A+B=90

B=90-A

$$\frac{\sqrt{\tan A \tan B + \tan A \cot B}}{\sin A \sec B}$$.

$$\frac{\sqrt{\tan A \tan(90-A) + \tan A \cot(90-A)}}{\sin A \sec (90-A)}$$.

$$\frac{\sqrt{\tan A \cot(A) + \tan A \tan(A)}}{\sin A \cosec (A)}$$.     

(Since We know that {$$\tan(90-A)=\cot A$$, $$\sec (90-A)=\cosec (A)$$,  $$\cot(90-A)= \tan A$$})

$$\frac{\sqrt{\tan A\frac{1} {\tan(A)} + \tan^2A }}{\sin A \cosec (A)}$$.

$$\frac{\sqrt{1 + \tan^2A }}{\sin A \frac{1}{\sin A}}$$.

$${\sqrt{1 + \tan^2A }}$$

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

$${ \sec A }$$.