pls tell!!!

Asked by nmishra320 | 22nd Jan, 2010, 05:40: PM

Expert Answer:

secθ + tanθ = p

(1+sinθ)/cosθ = p

Squaring both sides,

1+2sinθ+sin2θ = p2cos2θ

1+2sinθ+sin2θ = p2(1-sin2θ)

(p2+1)sin2θ+2sinθ+1-p2 = 0

sinθ = {-2±(4-4(p2+1)(1-p2)}/2(p2+1)

= {-1±(1-(p2+1)(1-p2))}/2(p2+1)

= {-1±(1-(p2+1)(1-p2))}/2(p2+1)

={-1±p2}/(p2+1)

Hence sinθ = (p2-1)/(p2+1)

Regards,

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Answered by  | 22nd Jan, 2010, 08:59: PM

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