CBSE Class 10 Answered
prove the following
Asked by narinder | 29 Nov, 2009, 02:18: PM
Expert Answer
1/ (sinθ - 2sin 3θ)/(2cos 3θ - cos θ) = sinθ(1 - 2sin2θ)/(cosθ(2cos2θ - 1))
= tanθ(sin2θ + cos2θ - 2sin2θ)/(2cos2θ-sin2θ-cos2θ)
=tanθ(cos2θ - 2sin2θ)/(cos2θ - 2sin2θ) = tanθ
2/ ( cosec A - sin A) (sec A - cos A) = (1/sinA - sinA)(1/cosA - cosA)
= ((1-sin2A)/sinA)((1-cos2A)/cosA) = (cos2A/sinA)(sin2A/cosA) = sinAcosA
= 1/(1/(sinAcosA)) = 1/((sin2A+cos2A)/(sinAcosA)) = 1/(sinA/cosA + cosA/sinA) = 1/(tanA + cotA)
Regards,
Team,
TopperLearning.
Answered by | 29 Nov, 2009, 05:46: PM
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