Question

√(sec^2A+cosec^2A)
$\sqrt{ { \sec(?) }^{2}a + { \csc(?) }^{2}a } = \tan(a) + \cot(a)$
=tanA+cotA

Answer 1

$\sqrt{sec^2 A + Cosec^2 A }= \sqrt{tan^2 A +1 + 1 + cot^2 A}\\\\=\sqrt{(tan\ A+ cot \: A)^2}= tan \: A + cot \: A$