# How do you simplify (sec^2x-1)/ sec^2x?

Nov 2, 2015

Simplify $\frac{{\sec}^{2} - 1}{\sec} ^ 2$

Ans: sin^2 x

#### Explanation:

Replace sec^2 = 1/(cos^2 x into the expression , we get:
$\frac{\frac{1}{{\cos}^{2} x - 1}}{\frac{1}{{\cos}^{2} x}} = \left(\frac{1 - {\cos}^{2} x}{\cos} ^ 2 x\right) \left(\frac{{\cos}^{2} x}{1}\right) =$

Since$\left(1 - {\cos}^{2} x\right) = {\sin}^{2} x$, therefore:

$\frac{{\sec}^{2} x - 1}{\sec} ^ 2 x = {\sin}^{2} x$

Nov 2, 2015

${\sin}^{2} x$

#### Explanation:

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

so $\text{ } {\sec}^{2} x - 1 = {\tan}^{2} x$

Giving $\text{ } \frac{{\tan}^{2} x}{{\sec}^{2} x}$

But $\text{ } {\sec}^{2} x = \frac{1}{{\cos}^{2} x}$

Giving $\text{ } \left({\tan}^{2} x\right) \left({\cos}^{2} x\right)$

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

Giving $\text{ } \left({\sin}^{2} x\right) \frac{{\cos}^{2} x}{{\cos}^{2} x} = {\sin}^{2} x$

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Clarification: By Tony B

Given:$\text{ } \frac{{\sec}^{2} x - 1}{\sec} ^ 2 x$

But $S e {c}^{2} x = 1 + {\tan}^{2} x$

Giving:$\text{ "(1+tan^2x-1)/(sec^2x)" "=" } {\tan}^{2} \frac{x}{\sec} ^ 2 x$

But $\text{ "tanx=sinx/cosx" and } \sec x = \frac{1}{\cos} x$

$\text{ "sin^2x/(cancel(cos^2x)) xxcancel(cos^2x)" "=" } {\sin}^{2} x$

Mar 5, 2016

$= {\sin}^{2} x$

#### Explanation:

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