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[Post New]2 Apr 2008 21:50:24 IST
    Subject: Re:if sinA ,sinB ,cosA r in G.P.then roots of x2+2xcotB+1=0 r always
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ankitagg (302)

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Olaaa!! Perrrfect answer. 50  [76 rates]
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x^2+2xcotB+1=0
x^2+2xcosB/sinB+1=0
(sinx)x^2+2xcosB+sinx=0

taking discrminant,
D=4cos^2B-4(sinB)(sinB)
 =4cos^2B-4sin^2B
 =4(1-2sin^2B)
 =4(1-2sinAcosA)   [sin^2B=sinAcosA ;by given G.P]
 =4(1-sin2A)
 =4-4sin2A

since,  -1<=sin2A<=1
therefore,D>=0
that implies roots are real.
 this reply: 5 points  (with 1 Olaaa!! Perrrfect answer.   in 1 votes )   [?]
 
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