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Answer 1 · 2016-06-21T07:40:22

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

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

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

54.#sin2xcosx-cos2xsinx#

#sin(2x-x)=sinx#

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

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

#=sinx/cosx=tanx#

56.#cos4x#
#=2cos^2 2x-1#
#=2(2cos^2x-1)^2-1#
#=4cos^4x-8cos^2x+2-1#
#:.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#