# How do you prove (tan u + cot u)(cos u + sin u) = csc u + sec u?

Sep 1, 2015

First, expand the Left Hand Side$\left(\text{LHS}\right)$

$\text{LHS} = \left(\tan u + \cot u\right) \left(\cos u + \sin u\right)$

$= \tan u \cos u + \tan u \sin u + \cot u \cos u + \cot u \sin u$

Recall that : $\tan u = \sin \frac{u}{\cos} u$
and $\text{ } \cot u = \cos \frac{u}{\sin} u$

$\implies \text{LHS} = \sin \frac{u}{\cancel{\cos u}} \cdot \cancel{\cos u} + \sin \frac{u}{\cos} u \cdot \sin u + \cos \frac{u}{\sin} u \cdot \cos u + \cos \frac{u}{\cancel{\sin u}} \cdot \cancel{\sin u}$

$= \sin u + {\sin}^{2} \frac{u}{\cos} u + {\cos}^{2} \frac{u}{\sin} u + \cos u$

$= \sin u + {\cos}^{2} \frac{u}{\sin} u + \cos u + {\sin}^{2} \frac{u}{\cos} u$

$= {\sin}^{2} \frac{u}{\sin} u + {\cos}^{2} \frac{u}{\sin} u + {\cos}^{2} \frac{u}{\cos} u + {\sin}^{2} \frac{u}{\cos} u$

$= \frac{{\sin}^{2} u + {\cos}^{2} u}{\sin} u + \frac{{\cos}^{2} u + {\sin}^{2} u}{\cos} u$

Recall again that : ${\sin}^{2} u + {\cos}^{2} u = 1$

$\implies \text{LHS} = \frac{1}{\sin} u + \frac{1}{\cos} u$

Recall that : $\frac{1}{\sin} u = \csc u$
and also $\text{ } \frac{1}{\cos} u = \sec u$

$\implies \text{LHS} = \textcolor{b l u e}{\csc u + \sec u}$