If atanθ=b, then acos2θ+bsin2θ=
a
b
-a
-b
Step1. Find the value of cos2θand sin2θ:
Since atanθ=b
tanθ=b/a….....(i)
cos2θ=(1–tan2θ)(1+tan2θ)
=1-b2a21+b2a2=(a2-b2)(a2+b2)
sin2θ=(2tanθ)(1+tan2θ)
=2ba1+b2a2=2ab(a2+b2)
Step2. Find the value of acos2θ+bsin2θ:
Now,
acos2θ+bsin2θ=a(a2-b2)(a2+b2)+b2ab(a2+b2)
=a(a2-b2)+2ab2(a2+b2)=a3-ab2+2ab2(a2+b2)=(a3+ab2)(a2+b2)=a(a2+b2)(a2+b2)=a
Hence, the correct option is (A).
pr.route sec^2theta+cosec^2theta=tantheta+cot theta