Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

$\dfrac{\text{tan}^2 θ}{\text{(sec θ - 1)}^2} = \dfrac{\text{1 + cos θ}}{\text{1 - cos θ}}$

The L.H.S. of the equation can be written as,

$\Rightarrow \dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)^2}{\left(\dfrac{1}{\cos\theta}-1\right)^2} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}}{\left(\dfrac{1-\cos\theta}{\cos\theta}\right)^2} \\[1em]$

Using $(\sin^2\theta=1-\cos^2\theta):$

$\Rightarrow \dfrac{\dfrac{1-\cos^2\theta}{\cos^2\theta}}{\dfrac{(1-\cos\theta)^2}{\cos^2\theta}} \\[1em] \Rightarrow \dfrac{1-\cos^2\theta}{(1-\cos\theta)^2} \\[1em] \Rightarrow \dfrac{(1-\cos\theta)(1+\cos\theta)}{(1-\cos\theta)^2} \\[1em] \Rightarrow \dfrac{\cancel{(1-\cos\theta)}(1+\cos\theta)}{(1-\cos\theta)^{\cancel{2}}} \\[1em] \Rightarrow \dfrac{1 + \cos\theta}{1 - \cos\theta} \\[1em]$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{tan}^2 θ}{\text{(sec θ - 1)}^2} = \dfrac{\text{1 + cos θ}}{\text{1 - cos θ}}$.