Question

If sec Θ = $\frac{2 \sqrt{3} }{3}$ and sin Θ < 0 , then find cot Θ + $\sqrt{2}$

Answer 1

$\sec x=\dfrac{1}{\cos x}\\\\\sec\theta=\dfrac{2\sqrt3}{3} \ \textgreater \ 0\ and\ \sin\theta \ \textless \ 0\to\theta\ is\ in\ the\ 4-th\ quadrant.\\\theta\in\left(\dfrac{3\pi}{2};\ 2\pi\right)\\\\\sec\theta=\dfrac{2\sqrt3}{3}\\\\\dfrac{1}{\cos\theta}=\dfrac{2\sqrt3}{3}\Rightarrow\cos\theta=\dfrac{3}{2\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}\\\\\cos\theta=\dfrac{3\sqrt3}{2\cdot3}\\\\\cos\theta=\dfrac{\sqrt3}{2}\to\theta=\dfrac{11\pi}{6}\\\\\cot\theta=\cot\frac{11\pi}{6}=\cot\frac{5\pi}{6}=-\sqrt3$

$Answer:\boxed{\sqrt2-\sqrt3}$