# How do you simplify (tanx + 1)^2 cosx?

Sep 6, 2016

$\sec x \left(1 + \sin 2 x\right)$

#### Explanation:

Let's begin by expanding the bracket.

${\left(\tan x + 1\right)}^{2} = {\tan}^{2} x + 2 \tan x + 1$

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

$\Rightarrow {\tan}^{2} x + 2 \tan x + 1 = \frac{{\sin}^{2} x}{{\cos}^{2} x} + \frac{2 \sin x}{\cos} x + 1$

substitute this back into the original.

That is $\left({\sin}^{2} \frac{x}{\cos} ^ 2 x + \frac{2 \sin x}{\cos} x + 1\right) \cos x$

distribute the bracket.

$\Rightarrow {\sin}^{2} \frac{x}{\cos} x + 2 \sin x + \cos x$

express each term with a common denominator of cosx

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

$= \frac{{\sin}^{2} x + 2 \sin x \cos x + {\cos}^{2} x}{\cos x} \ldots \ldots . . \left(A\right)$

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

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin 2 x = 2 \sin x \cos x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Returning to (A)

$\Rightarrow \frac{1 + \sin 2 x}{\cos} x = \frac{1}{\cos} x + \frac{\sin 2 x}{\cos} x$

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

$\Rightarrow \frac{1}{\cos} x + \frac{\sin 2 x}{\cos} x = \sec x + \sec x \sin 2 x$

$\Rightarrow {\left(\tan x + 1\right)}^{2} \cos x = \sec x \left(1 + \sin 2 x\right)$