IF SecA+TanA=p P.T SinA=p2[Square]-1/p2[Square]+1

We know that sec² A - tan² A =1 and it is given that
sec A + tan A =p

dividing equation 1 with 2, we get

sec A - tan A = 1/p
Also sec A + tan A = p

Adding and subtracting these equations, we get
sec A = (p + 1/p )/2
& tan A = (p-1/p)/2

Thus, sin A = (sin A/ cos A) / (1/cos A) = tan A / sec A =(p² - 1) / (p²+ 1)