1.If tanθ−2secθ=3\tan\theta - \sqrt{2}\sec\theta = \sqrt{3}tanθ−2secθ=3, then the general solution of θ\thetaθ isa.nπ+(−1)nπ4+π3n\pi + (-1)^n\dfrac{\pi}{4} + \dfrac{\pi}{3}nπ+(−1)n4π+3πb.nπ+(−1)n3π4n\pi + (-1)^n\dfrac{3\pi}{4}nπ+(−1)n43πc.nπ+(−1)nπ3+π4n\pi + (-1)^n\dfrac{\pi}{3} + \dfrac{\pi}{4}nπ+(−1)n3π+4πd.nπ+(−1)nπ4+π3n\pi + (-1)^n\dfrac{\pi}{4} + \dfrac{\pi}{3}nπ+(−1)n4π+3πLogin to continueOnly logged in users canattempt or see the solution.
