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[Post New]16 Apr 2008 20:18:42 IST
    Subject: Re:please answer
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sboosy (3046)

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Olaaa!! Perrrfect answer. 536  [719 rates]
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4x^2-16x+15<0 \\ \\ \Rightarrow (2x-3)(2x-5)<0 \\ \\ \mbox{Only integral solution is} \ x=2 \ \Rightarrow \cot(\theta_1)=2 \ \Rightarrow \sin(\theta_1)=\frac{1}{\sqrt{5}} \\ \\ \mbox{Bisector of 1st quad means slope} \ =\tan(45) =1 = \sin(\theta_2) \\ \\ \sin(\theta_1+\theta_2)*\sin(\theta_1-\theta_2) = \sin^2(\theta_1) - \sin^2(\theta_2) = \frac{1}{5}-1 = \frac{-4}{5}
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