Question

Root 1-sinA/1+sinA=secA-tanA

Answer 1

Whole root of ( 1+ sinA) / (1 - sinA ) = whole root of (1+sinA)(1+sinA)/(1-sinA)(1+sinA) (multiplying and dividing by 1+sinA within the root, or in other words, by conjugation) 
= whole root of (1+sinA)^2/(1-sin^2A) 
= whole root of (1+sinA)^2/cos^2A ( since 1-sin^2A=cos^2A) 
= whole root of [(1+sinA)/ cosA]^2 
= (1+sinA)/cosA (the root gets cancelled with the square of the fraction) 
= (1/cosA)+(sinA/cosA) (splitting the denominator among the 2 numerator elements) 
= secA+tanA