Prove the following :

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

To prove,

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

We know that,

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

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

$\Rightarrow \dfrac{\text{cos θ}}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

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

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