# How do you determine graphically and analytically whether y_1=sec^2x-1 is equivalent to y_2=tan^2x?

Feb 24, 2018

See below.

#### Explanation:

Identities:

1) $\textcolor{w h i t e}{88} \textcolor{red}{\boldsymbol{{\sin}^{2} x + {\cos}^{2} x = 1}}$

2)$\textcolor{w h i t e}{88} \textcolor{red}{\boldsymbol{{\sec}^{2} x = \frac{1}{\cos} ^ 2 x}}$

3)$\textcolor{w h i t e}{88} \textcolor{red}{\boldsymbol{{\tan}^{2} x = {\sin}^{2} \frac{x}{\cos} ^ 2 x}}$

${y}_{1} = {y}_{2} \implies {\sec}^{2} x - 1 = {\tan}^{2} x$

LHS

By identity 2.

$\frac{1}{\cos} ^ 2 x - 1$

$\frac{1 - {\cos}^{2} x}{\cos} ^ 2 x$

From identity 1:

$1 - {\cos}^{2} x = {\sin}^{2} x$

Hence:

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

By identity 3:

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

So:

$L H S \equiv R H S$

This is an identity, so they are equivalent.

Graphically:

Plot both $\boldsymbol{{\sec}^{2} x - 1}$ and $\boldsymbol{{\tan}^{2} x}$. As you can see their graphs are identical, as we would expect.

$\boldsymbol{y = {\sec}^{2} x - 1}$

graph{y=(tan(x))^2 [-11.25, 11.25, -5.625, 5.625]}

$\boldsymbol{y = {\tan}^{2} x}$
graph{y=(sec(x))^2-1 [-11.25, 11.25, -5.625, 5.625]}