C-09-Q190mediumsingle_mcqSolve: tan2θ−(1+3)tanθ+3=0\tan^2\theta - (1 + \sqrt{3})\tan\theta + \sqrt{3} = 0tan2θ−(1+3)tanθ+3=0. θ=?\theta = ?θ=?aonly 45∘45^\circ45∘bonly 60∘60^\circ60∘c45∘45^\circ45∘ or 60∘60^\circ60∘d30∘30^\circ30∘ব্যাখ্যাThe quadratic factors as (tanθ−1)(tanθ−3)=0(\tan\theta-1)(\tan\theta-\sqrt3)=0(tanθ−1)(tanθ−3)=0, giving tanθ=1\tan\theta=1tanθ=1 or tanθ=3\tan\theta=\sqrt3tanθ=3. These correspond to θ=45∘\theta=45^\circθ=45∘ or 60∘60^\circ60∘.