Prove that :

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

Solving L.H.S. of above equation,

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

Since, L.H.S. = R.H.S.

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