prove the following

Asked by narinder | 29th 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  | 29th Nov, 2009, 05:46: PM

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