Question

How do you show that `(tan^2x + 1)/(1 - tan^2x) = sec^2x`?

Answer 1

This is true for all values of x, because the attempt to solve resulted in a proof.

Explanation:

Given: `(tan^2(x) + 1)/(1 - tan^2(x)) = sec(2x)`

Write every occurrence of `tan^2(x)" as " sin^2(x)/cos^2(x)`

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

Multiply the left side by 1 in the form of `cos^2(x)/cos^2(x)`

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

Simplify the left side:

`(sin^2(x)+cos^2(x))/(cos^2(x) - sin^2(x)) = sec(2x)`

The numerator is known to be equal to 1:

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

The denominator is known to be equal to `cos(2x)`

`1/cos(2x) = sec(2x)`

This is a well known identity and, therefore, is true for every value of x.

`sec(2x) = sec(2x)`

Answer 2

We start by rewriting everything in sine and cosine. We do this using the identities `sectheta = 1/costheta` and `tantheta = sintheta/costheta`.

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

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

Use `cos(2x) = cos^2x - sin^2x` and `sin^2x + cos^2x = 1`.

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

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

Hopefully this helps!

Answer 3

Use Pythagorean identity `tan^2 x + 1 = sec^2 x` and double angle identity `cos2x = cos^2x-sin^2x`.

Explanation:

Using the Pythagorean identity `tan^2 x + 1 = sec^2 x`, we get

`(tan^2 x + 1)/(1 - tan^2 x) = (sec^2 x)/(1 - tan^2x)`

`color(white)((tan^2 x + 1)/(1 - tan^2 x)) = 1/(cos^2x(1 - tan^2x))` (since `sec theta = 1/cos theta`)

`color(white)((tan^2 x + 1)/(1 - tan^2 x)) = 1/(cos^2x - sin^2x)"      "` (distributing the `cos^2x`)

`color(white)((tan^2 x + 1)/(1 - tan^2 x)) = 1/(cos2x)"              "` (by double angle identity)

`color(white)((tan^2 x + 1)/(1 - tan^2 x)) = sec2x"                "` (by reciprocal identity)