`cos^4A-sin^4A+1=2cos^2A`

First, group the first two terms and factor it.

`(cos^4A-sin^4A)+1=2cos^2A`

`(cos^2A-sin^2A)(cos^2A+sin^2A)+1=2cos^2A`

Then, apply the Pythagorean identity which is `cos^2theta + sin^2theta=1` .

`(cos^2A-sin^2A)(1)+1=2cos^2A`

`cos^2A-sin^2A + 1=2cos^2A`

Then, group the second and last term at the left side of the equation.

`cos^2A+(-sin^2A+1)=2cos^2A`

`cos^2A+(1-sin^2A)=2cos^2A`

To simplify the expression inside the...

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`cos^4A-sin^4A+1=2cos^2A`

First, group the first two terms and factor it.

`(cos^4A-sin^4A)+1=2cos^2A`

`(cos^2A-sin^2A)(cos^2A+sin^2A)+1=2cos^2A`

Then, apply the Pythagorean identity which is `cos^2theta + sin^2theta=1` .

`(cos^2A-sin^2A)(1)+1=2cos^2A`

`cos^2A-sin^2A + 1=2cos^2A`

Then, group the second and last term at the left side of the equation.

`cos^2A+(-sin^2A+1)=2cos^2A`

`cos^2A+(1-sin^2A)=2cos^2A`

To simplify the expression inside the parenthesis, apply the Pythagorean identity again.

`cos^2A+cos^2A=2cos^2A`

`2cos^2A=2cos^2A`

**Since left side simplifies to `2cos^2A` which is the same term with the right side, hence it proves that the given equation is an identity.**