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

Apr 13, 2018

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

#### Explanation:

First, convert all of the trigonometric functions to $\sin \left(x\right)$ and $\cos \left(x\right)$:

$\frac{{\sec}^{2} \left(x\right) - 1}{\sin} ^ 2 \left(x\right)$

$= \frac{\frac{1}{\cos} ^ 2 \left(x\right) - 1}{\sin} ^ 2 \left(x\right)$

$= \frac{\frac{1 - {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right)}{\sin} ^ 2 \left(x\right)$

Use the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$:

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

Canceling out the ${\sin}^{2} \left(x\right)$ present in both the numerator and the denominator:

$= \frac{1}{\cos} ^ 2 \left(x\right)$

$= {\sec}^{2} \left(x\right)$

Apr 13, 2018

The answer is ${\sec}^{2} x$.

#### Explanation:

We know that,
${\sec}^{2} x - 1 = {\tan}^{2} x$

Therefore,$\frac{{\sec}^{2} x - 1}{\sin} ^ 2 x$
=${\tan}^{2} \frac{x}{\sin} ^ 2 x$
=${\sin}^{2} \frac{x}{\cos} ^ 2 x \cdot \frac{1}{\sin} ^ 2 x$
=$\frac{1}{\cos} ^ 2 x$

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

Apr 13, 2018

${\sec}^{2} x$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)secx=1/cosx

•color(white)(x)sin^2x+cos^2x=1

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

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

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

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

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