Question

How to prove the identity ? `1- (tan^2x/(1+tan^2x)) = cos^2x`

Answer 1

`1- (tan^2x/(1+tan^2x)) = cos^2x` `color(white)(ddddwwwdd` `["as " color(red)(tanx = sinx/cosx)]`

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

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

`1- (sin^2x/1) = cos^2x` `color(white)(ddddwwwdd` `["as " color(red)(cos^2x+sin^2x=1)]`

`1- sin^2x = cos^2x`

`cos^2x=cos^2x` `color(white)(ddddwwwwwwwwdd` `["as " color(red)(cos^2x=1-sin^2x)]`

Hence Proved !

Answer 2

1. `1+tan^2x = sec^2x`
2. `1 - (tan^2x/sec^2x) = cos^2x`
3. `tan^2x/sec^2x = sin^2x`
4. `1 - sin^2x = cos^2x`

Explanation:

Line 1 is a common Pythagorean identity. Substituting gives line 2. Simplifying the complex fraction in 2 into sin and cos functions gives line 3. Line 4 is also a common Pythagorean identity.

Answer 3

Here is how I proved the identity:

Explanation:

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

To solve this, we will use a bunch of trigonometric identities.

First, we know that `tan^2x = sin^2x/cos^2x` from quotient identities.
We also know that `1 + tan^2x = sec^2x` from the pythagorean identities.

From these identities, we can rewrite the equation as:
`1 - (sin^2x/cos^2x)/(sec^2x) = cos^2x`

We also know that `sec^2x = 1/cos^2x` from reciprocal functions:
`1 - (sin^2x/cos^2x)/(1/cos^2x) = cos^2x`

Let's rewrite the division part with `-:` instead of `/` so it is easier to read:
`1 - (sin^2x/cos^2x) -: (1/cos^2x) = cos^2x`

We know that dividing something is the same as multiplying it by the reciprocal, or `1` over the number:
`1 - (sin^2x/cos^2x) * (cos^2x/1) = cos^2x`

Since `cos^2x` is being both divided and multiplied, we can cross both of them out:
`1 - (sin^2x/cancel(cos^2x)) * (cancel(cos^2x)/1) = cos^2x`

And now the cleaned up version looks like this:
`1 - sin^2x = cos^2x`

Finally, we know that `1-sin^2x = cos^2x` from the pythagorean identities:
`cos^2x = cos^2x`

Hope this helps!