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Q.

If tanθ=ba  then a cos2θ+bsin2θ=

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a

a

b

b

c

ba

d

ab

answer is B.

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Detailed Solution

acos2θ+bsin2θ=a[1tan2θ1+tan2θ]+b[2tanθ1+tan2θ]

a[a2b2a2+b2]+b[2baa2+b2]

=(a(a2b2)+b(2ab)a2+b2)

=a3ab2+2ab2a2+b2

=a3+ab2a2+b2

=a(a2+b2)a2+b2

=a

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If tanθ=ba  then a cos2θ+bsin2θ=