CBSE Class 10 Answered
pls tell!!!
Asked by nmishra320 | 22 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,
Team,
TopperLearning.
Answered by | 22 Jan, 2010, 08:59: PM
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