1.If 2tan2θ=sec2θ2\tan^2\theta = \sec^2\theta2tan2θ=sec2θ, then the general solution of θ\thetaθ isa.nπ+π4n\pi + \dfrac{\pi}{4}nπ+4πb.nπ−π4n\pi - \dfrac{\pi}{4}nπ−4πc.nπ±π4n\pi \pm \dfrac{\pi}{4}nπ±4πd.2nπ+π42n\pi + \dfrac{\pi}{4}2nπ+4πLogin to continueOnly logged in users canattempt or see the solution.
