Prove the following :

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

To prove,

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

We know that,

sin (90 - θ) = cos θ and tan (90 - θ) = cot θ

Solving L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{tan θ}}{\text{\text{cot θ}}} + \dfrac{\text{cos θ}}{\text{cos θ}} \\[1em] \Rightarrow \dfrac{\text{tan θ}}{\dfrac{1}{\text{tan θ}}} + 1 \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + 1 \\[1em] \Rightarrow \text{sec}^2 \text{ θ}.$

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

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