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Question

If m=cosAsinA and n=cosA+sinA; show that m2+n2m2n2=12secA.cosecA=(cotA+tanA)2.

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Solution

m=cosA-sinA , m=cosA+sinA.
m2+n2m2n2=(cosAsinA)2(cosA+sinA)2(cosAsinA)2(cosA+sinA)2
=cos2A+sin2A2cosAsinA+cos2A+sin2A+2cosAsinA(cosAsinA+cosA+sinA)(cosAsinAcosAsinA)
=1+12cosA(2sina)
=24cosAsinA
=12secAcosA.
=12cosAsinA
=12sin2A[sin2θ=2sinθcosθ]
=12tan41+tan2A[sin2θ=2tan2θ1+tan2θ]
=(1+tan2A)2tanA
=12(1tana+tan2AtanA)
=12(cotA+tanA)


1216008_1388188_ans_af2bb3261cea42aaacdfe4ee090fae99.JPG

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