let's see
cos(θ)sin(θ) > 0, namely their product is positive.
well, that can only happen is both are the same sign, either + or -, and that only happens in the III and I Quadrants.
sec(θ) < 0, namely secant is negative.
recall sec(θ) = 1/cos(θ)
so the secant can only be negative if the cosine is negative as well.
so both cosine and sine must be the same signs, and cosine must be negative, so both are negative, that means the angle is in the III Quadrant.
[tex]\bf sin^2(\theta )=3cos^2(\theta )\implies sin^2(\theta )=3[1-sin^2(\theta )]\implies sin^2(\theta )=3-3sin^2(\theta ) \\\\\\ 4sin^2(\theta )=3\implies sin^2(\theta )=\cfrac{3}{4}\implies sin(\theta )=\pm\sqrt{\cfrac{3}{4}}\implies sin(\theta )=\pm\cfrac{\sqrt{3}}{\sqrt{4}} \\\\\\ sin(\theta )=\pm\cfrac{\sqrt{3}}{2}\implies sin(\theta )=-\cfrac{\sqrt{3}}{2}\qquad \leftarrow \textit{III Quadrant}\to \qquad \cfrac{4\pi }{3}[/tex]
now is just a matter of getting those values from our hand dandy Unit Circle
[tex]\bf cos(\theta )=-\cfrac{1}{2}\qquad \qquad tan(\theta )=\sqrt{3}\qquad \qquad sec(\theta )=-2 \\\\\\ cot(\theta )=\cfrac{1}{\sqrt{3}}\implies \stackrel{\textit{rationalizing the denominator}}{cot(\theta )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies cot(\theta )=\cfrac{\sqrt{3}}{3}} \\\\\\ csc(\theta )=-\cfrac{2}{\sqrt{3}}\stackrel{\textit{rationalizing the denominator}}{csc(\theta )=-\cfrac{2}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies csc(\theta )=-\cfrac{2\sqrt{3}}{3}}[/tex]
