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!