Question #abcc5

1 Answer
Jun 21, 2016

52.(1-cos^2theta)(1+cot^2theta)(1cos2θ)(1+cot2θ)

=sin^2theta*csc^2theta=sin2θcsc2θ
=1=1

53.cos(2x-3y)cos(2x3y)
=cos2xcos3y+sin2xsin3y=cos2xcos3y+sin2xsin3y

54.sin2xcosx-cos2xsinxsin2xcosxcos2xsinx

sin(2x-x)=sinxsin(2xx)=sinx

55.(1-cos2x)/(sin2x)1cos2xsin2x

=(2sin^2x)/(2sinxcosx)=2sin2x2sinxcosx

=sinx/cosx=tanx=sinxcosx=tanx

56.cos4xcos4x
=2cos^2 2x-1=2cos22x1
=2(2cos^2x-1)^2-1=2(2cos2x1)21
=4cos^4x-8cos^2x+2-1=4cos4x8cos2x+21
:.cos4x=4cos^4x-8cos^2x+1..(1)

Given

cos4x=pcos^4x+qcos^x+r...(2)

Comparing (1) and (2)

we have
p=4,q=-8,r=1

57.
costheta/(1-sintheta)

=(cos^2(theta/2)-sin^2(theta/2))/(cos^2(theta/2)+sin^2(theta/2)-2sin(theta/2)cos(theta/2))

=(cos^2(theta/2)-sin^2(theta/2))/(cos(theta/2)-sin(theta/2))^2

=(cos(theta/2)+sin(theta/2))/(cos(theta/2)-sin(theta/2))

Dividing numerator and denominator by cos(theta/2)
we get
=(1+tan(theta/2))/(1-tan(theta/2))

=(t+1)/(1-t)

Putting tan(theta/2)=t