# How do you simplify the expression cott(tant+cott)?

Sep 12, 2016

${\csc}^{2} t$

#### Explanation:

We can begin by using the $\textcolor{b l u e}{\text{trigonometric identities}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\tan t = \frac{\sin t}{\cos t} \text{ and } \cot t = \frac{\cos t}{\sin t}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The expression may now be written as.

$\frac{\cos t}{\sin t} \left(\frac{\sin t}{\cos t} + \frac{\cos t}{\sin t}\right)$

and distributing the bracket gives.

$1 + \frac{{\cos}^{2} t}{{\sin}^{2} t} = \frac{{\sin}^{2} t}{{\sin}^{2} t} + \frac{{\cos}^{2} t}{{\sin}^{2} t}$

both fractions have a common denominator so adding gives.

$\frac{{\sin}^{2} t + {\cos}^{2} t}{{\sin}^{2} t}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} t + {\cos}^{2} t = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\csc t = \frac{1}{\sin t}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \frac{{\sin}^{t} + {\cos}^{2} t}{{\sin}^{2} t} = \frac{1}{{\sin}^{2} t} = {\csc}^{2} t$

Thus $\cot t \left(\tan t + \cot t\right) \text{ simplifies to} {\csc}^{2} t$