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

3 Answers
Apr 3, 2018

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 !

Apr 3, 2018
  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.

Apr 3, 2018

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!