Question

prove that :--
√sec^2 A + cosec^2 = tan A + cot A​

Answer 1

LHS

$\sqrt{sec^2A+cosec^2A}$

=$\sqrt{1+tan^2A+1+cot^2A}$   ( by identity  sec²A-tan²A=1 and cosec²A-cot²A=1)

=$\sqrt{tan^2A+cot^2A+2}$

=$\sqrt{tan^2A+cot^2A+2tanA.cotA}$   ( as $tanA*cotA=1$)

=$\sqrt{(tanA+cotA)^2}$ [ as (a+b)²=a²+b²+2ab)]

=$tanA+cotA$

LHS=RHS

Hence proved