Simplify the following :

$\dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30°$

We know that,

sin (90° - θ) = cos θ, cos (90° - θ) = sin θ and sec (90° - θ) = cosec θ.

Substituting values in equation, we get :

$\Rightarrow \dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30° \\[1em] \Rightarrow \dfrac{\text{cos θ}}{\text{cos θ}} + \dfrac{\text{\text{sin θ}}}{\text{cosec θ}} - \text{3 tan}^2 30° \\[1em] \Rightarrow 1 + \dfrac{\text{sin θ}}{\dfrac{1}{\text{sin θ}}} -3 \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] = 1 + \text{sin}^2 θ - 3 \times \dfrac{1}{3} \\[1em] = 1 - 1 + \text{sin}^2 θ\\[1em] = \text{sin}^2 θ.$

Hence, $\dfrac{\text{cos θ}}{\text{sin (90° - θ)}} + \dfrac{\text{\text{cos (90° - θ)}}}{\text{sec (90° - θ)}} - \text{3 tan}^2 30°$ = sin² θ.