If sec A + tan A = p, then the value of sin A is?

secA+tanA =p

1+sinA/cosA = p

1+sinA = pcosA

squaring both sides

1+ sin²A + 2sinA = p²cos²A           (cos²A = 1-sin²A)

sin²A(1+p²) + 2sinA + 1-p²  = 0

this is a quadratic eq and roots of eq are

sinA = [-b+(-)(b²-4ac)^1/2]/2a

       =-1 ,  (p²-1/p²+1)

sinA = -1 cannot be the root of the eq because it doesnot satisfies the given eqpression so

the only root is

sinA = p²-1/p²+1