1.If tan2θ−(1+3)tanθ+3=0\tan^2\theta - (1 + \sqrt{3})\tan\theta + \sqrt{3} = 0tan2θ−(1+3)tanθ+3=0, then the general value of θ\thetaθ isa.nπ+π4, nπ+π3n\pi + \dfrac{\pi}{4},\; n\pi + \dfrac{\pi}{3}nπ+4π,nπ+3πb.nπ−π4, nπ+π3n\pi - \dfrac{\pi}{4},\; n\pi + \dfrac{\pi}{3}nπ−4π,nπ+3πc.nπ+π4, nπ−π3n\pi + \dfrac{\pi}{4},\; n\pi - \dfrac{\pi}{3}nπ+4π,nπ−3πd.nπ−π4, nπ−π3n\pi - \dfrac{\pi}{4},\; n\pi - \dfrac{\pi}{3}nπ−4π,nπ−3πLogin to continueOnly logged in users canattempt or see the solution.
