# How do you verify (tan^3(x) - 1) / (tan(x) - 1) = sec^2(x) + tan(x)?

$\frac{{\tan}^{3} \left(x\right) - 1}{\tan \left(x\right) - 1}$
$\text{Since } {a}^{3} - {b}^{3} = \left(a - b\right) \cdot \left({a}^{2} + a \cdot b + {b}^{2}\right)$, we may write the LHS
$= \frac{\cancel{\left(\tan \left(x\right) - 1\right)} \cdot \left({\tan}^{2} \left(x\right) + \tan \left(x\right) + 1\right)}{\cancel{\left(\tan \left(x\right) - 1\right)}}$
$= \cancel{{\tan}^{2} \left(x\right)} + \tan \left(x\right) + S e {c}^{2} \left(x\right) - \cancel{{\tan}^{2} \left(x\right)}$ $\left[\text{Since } S e {c}^{2} \left(x\right) - {\tan}^{2} \left(x\right) = 1\right]$
$= S e {c}^{2} \left(x\right) + \tan \left(x\right) =$RHS Proved.