If $\sqrt{2}$ tan 2θ = $\sqrt{6}$ and 0° < 2θ < 90°, find the value of

sin θ + $\sqrt{3}$ cos θ - 2 tan² θ.

Given,

$\phantom{\Rightarrow}\sqrt{2}$ tan 2θ = $\sqrt{6}$

⇒ tan 2θ = $\dfrac{\sqrt{6}}{\sqrt{2}}$

⇒ tan 2θ = $\sqrt{3}$

⇒ tan 2θ = tan 60°

⇒ 2θ = 60°

⇒ θ = $\dfrac{60°}{2}$

⇒ θ = 30°.

Substituting value of θ in sin θ + $\sqrt{3}$ cos θ - 2 tan² θ, we get :

⇒ sin 30° + $\sqrt{3}$ cos 30° - 2 tan² 30°

$\Rightarrow \dfrac{1}{2} + \sqrt{3} \times \dfrac{\sqrt{3}}{2} - 2 \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} - \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3 + 9 - 4}{6} \\[1em] \Rightarrow \dfrac{8}{6} \\[1em] \Rightarrow \dfrac{4}{3}.$

Hence, sin θ + $\sqrt{3}$ cos θ - 2 tan² θ = $\dfrac{4}{3}.$