# How do you simplify the expression tan^2x/(secx+1)?

Sep 1, 2016

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

$\Rightarrow {\tan}^{2} \frac{x}{\sec x + 1} = \sec x - 1 , \mathmr{and} , = \frac{1 - \cos x}{\cos} x$,

Sep 1, 2016

$\sec x - 1$

#### Explanation:

Use the $\textcolor{b l u e}{\text{trigonometric identity}}$ for ${\tan}^{2} x$

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

$\Rightarrow \frac{{\tan}^{2} x}{\sec x + 1} = \frac{{\sec}^{2} x - 1}{\sec x + 1} \ldots \ldots . . \left(A\right)$

Note that ${\sec}^{2} x - 1 \text{ is a difference of squares}$ and in general, factorises as.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${\left(\sec x\right)}^{2} = {\sec}^{2} x \text{ and " (1)^2=1rArra=secx" and } b = 1$

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

substitute into (A)

$\Rightarrow \frac{{\sec}^{2} x - 1}{\sec x + 1} = \frac{\left(\sec x - 1\right) \cancel{\left(\sec x + 1\right)}}{\cancel{\sec x + 1}}$

$\Rightarrow \frac{{\tan}^{2} x}{\sec x + 1} = \sec x - 1$