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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
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