Question

A+B=90 then prove that √(tanA tan B + tan A cot B)/(sinA SecB) = secA

Answer 1

√(tanA tan B + tan A cot B)/(sinA secB)

∵ tanB = cot(90-B); cotB = tan(90-B); secB = cosec(90-B) & 90-B=A

∴√(tanA cotA + tan A tanA)/(sinA cosecA) 

=√(1+tan²A)/1

=√sec²A/1

=secA/1 = secA .................. Hence Proved

HOPE IT HELPS................... :)