If tanθ=b/a than the value of acos2θ+bsin2θ

Steps to find the value of acos2θ+bsin2θ

We know that cos2θ=(1-tan^2θ)/(1+tan^2θ)

and sin2θ=2tanθ/(1-tan^2θ)

=a.(1-tan^2θ)/(1+tan^2θ)+b.2tanθ/(1-tan^2θ)

Put tanθ=b/a

=a.{1-(b/a)^2}/{1+(b/a)^2}+b.2(b/a)/{1+(b/a)^2}

=(a^3-ab^2)/(a^2+b^2)+2ab^2/(a^2+b^2)

=(a^3-ab^2+2ab^2)/(a^2+b^2)

=(a^3+ab^2)/(a^2+b^2)

=a(a^2+b^2)/(a^2+b^2)

=a

Hence If tanθ=b/a than the value of acos2θ+bsin2θ is a.

Detail Information:-

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