MathematicsHard64×since 2004Q4021If ∫dθcos2θ(tan2θ+sec2θ)=λtanθ+2loge∣f(θ)∣+C\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C∫cos2θ(tan2θ+sec2θ)dθ=λtanθ+2loge∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ\lambdaλ, ƒ(θ\thetaθ)) is equal to :A(–1, 1 – tanθ\thetaθ)B(1, 1 + tanθ\thetaθ)C(–1, 1 + tanθ\thetaθ)D(1, 1 – tanθ\thetaθ)Check answerSkip