Prove the following :

$\dfrac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \dfrac{1}{\text{cos \text{ θ}}}$

To prove,

$\dfrac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \dfrac{1}{\text{cos \text{ θ}}}$

Solving LHS of the above equation,

$\phantom{=} \dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} \\[1em] = \dfrac{\text{sin}^2 \text{ θ} + \text{cos}^2 θ}{\text{cos θ}} \\[1em] = \dfrac{1}{\text{cos θ}} \space [\because \text{sin}^2 \text{ θ} + \text{ cos}^2 \text{ θ} = 1]$

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

Hence, proved that $\dfrac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} = \dfrac{1}{\text{cos θ}}.$