Here we have to prove that : (sin A)^4 + 2 (cos A)^2 - (cos A)^4 = 1.

We use the relation that ( sin A)^2 + (cos A)^2 = 1

(sin A)^4 + 2 (cos A)^2 - (cos A)^4

=> [(sin A)^2]^2 + 2 (cos A)^2 - [(cos A)^2]^2

=> [(sin A)^2]^2 + 2*[1 - (sin A)^2] - [(cos A)^2]^2

=> [(sin A)^2]^2 + [2 - 2*(sin A)^2] - [(cos A)^2]^2

=> [(sin A)^2]^2 - [(cos A)^2]^2 + [2 - 2*(sin A)^2]

using a^2 - b^2 = ( a - b)( a + b)

=> [(sin A)^2 - (cos A)^2][(sin A)^2 + (cos A)^2]+ [2 - 2*(sin A)^2]

=> [(sin A)^2 - (cos A)^2]*1 + [2 - 2*(sin A)^2]

=> (sin A)^2 - (cos A)^2 + 2 - 2*(sin A)^2

=> - (cos A)^2 + 2 - (sin A)^2

=> - [ (cos A)^2 + (sin A)^2] + 2

=> -1 + 2

=> 1

**Therefore (sin A)^4 + 2 (cos A)^2 - (cos A)^4 = 1.**

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

First, we will rearrange terms.

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

Now we know that: (a^2 - b^2) = (a-b)(a+b)

==> sin^4 A -cos^4 A = (sin^2 A - cos^2 A)(sin^2 A+cos^2 A)

But we know that sin^2 A + cos^2 A = 1

==> sin^4 A - cos^4 A = ( sin^2 A - cos^2 A)

Now we will substitute:

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

We will combine like terms.

==> sin^2 A + cos^2 A = 1

==> 1 = 1

**==> sin^4 A + 2cos^2 A - cos^4 A = 1............q.e.d**

L:H:S ≡ sin ⁴A + 2cos²A - cos ⁴A

= (sin²A)² - (cos²A)² + 2cos²A

*⇒ use x² - Y² = (x+y)(x-y)*

= (sin²A + cos²A)(sin²A - cos²A) + 2cos²A

*⇒ use sin²A + cos²A = 1*

= (sin²A - cos²A) + 2cos²A

= (1 - cos²A - cos²A) + 2cos²A

= 1 - 2cos²A + 2cos²A

= 1

Hence L:H:S ≡ R:H:S

To prove sin^4A + 2cos^2A - cos^4A = 1.

Consider the term sin^4x = (1-cos^2x)^2.

(1-cos^2x)^2 = 1-2cos^2x +cos^4x.

Therefore we replace sin^4x by (1-cos^2x)^2 = 1-2cos^2x+cos^4x. in the left of the given identity.

So LHS : sin^4A + 2cos^2A - cos^4A = (1-2cos^2x+cos^4x)+2cos^2x+cos^4x.

sin^4A + 2cos^2A - cos^4A = 1-2cos^2x+cos^4x+2cos^2x+cos^4x = 1.

Therefore sin^4A + 2cos^2A - cos^4A = 1.