Question

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

Answer 1

`(sec^2(x)-1)/sin^2(x)=sec^2(x)`

Explanation:

First, convert all of the trigonometric functions to `sin(x)` and `cos(x)`:

`(sec^2(x)-1)/sin^2(x)`

`=(1/cos^2(x)-1)/sin^2(x)`

`=((1-cos^2(x))/cos^2(x))/sin^2(x)`

Use the identity `sin^2(x)+cos^2(x)=1`:

`=(sin^2(x)/cos^2(x))/sin^2(x)`

Canceling out the `sin^2(x)` present in both the numerator and the denominator:

`=1/cos^2(x)`

`=sec^2(x)`

Answer 2

The answer is `sec^2x`.

Explanation:

We know that,
`sec^2x-1=tan^2x`

Therefore,`(sec^2x-1)/sin^2x`
=`tan^2x/sin^2x`
=`sin^2x/cos^2x*1/sin^2x`
=`1/cos^2x`

=`sec^2x`

Answer 3

`sec^2x`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)secx=1/cosx`

`•color(white)(x)sin^2x+cos^2x=1`

`rArr(1/cos^2x-cos^2x/cos^2x)/sin^2x`

`=((1-cos^2x)/cos^2x)/sin^2x`

`=(sin^2x/cos^2x)/sin^2x`

`=cancel(sin^2x)/cos^2x xx1/cancel(sin^2x)`

`=1/cos^2x=sec^2x`