Simplify the following :

$\dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}}.$

We know that,

sin (90° - θ) = cos θ
cos (90° - θ) = sin θ
sec (90° - θ) = cosec θ
cosec (90° - θ) = cot θ
cot (90° - θ) = tan θ.

Substituting values in equation, we get :

$\Rightarrow \dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \dfrac{\text{sec θ cos θ tan θ}}{\text{sin θ cosec θ tan θ}} + \dfrac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} \times \text{cos θ} \times \text{tan θ}}{\text{sin θ} \times \dfrac{1}{\text{sin θ}} \times \text{tan θ}} + 1 \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

Hence, $\dfrac{\text{cosec (90° - θ) sin (90° - θ) cot(90° - θ)}}{\text{cos (90° - θ) sec (90° - θ) tan θ}} + \dfrac{\text{cot θ}}{\text{tan (90° - θ)}}$ = 2.