Prove that \[{\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A.{\cos ^2}A\]

Hint:
The most important two Trigonometric functions are given in the given problem. We are to prove that the left hand side of the problem is equal to the right hand side.
We are to use the property.
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Given that the angle is the same for both the T-functions.
Also, the identity written as follows
\[{\left( {a + b} \right)^2} + {a^2} + {b^2} + 2.a.b\] is going to be used in this problem.

Complete step by step solution:
We are to prove that
\[{\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A.{\cos ^2}A\]
As we can notice, the angle in this problem is the same for both the T-functions i.e. A
Step – 1 :
Consider the property
\[{\sin ^2}A + {\cos ^2}A = 1\]
Step – 2 :
Taking squares on both the sides of the identity
\[{\left( {{{\sin }^2}A + {{\cos }^2}} \right)^2} = {\left( 1 \right)^2}\]
Step – 3 :
On solving the identity, we get;
\[{\left( {{{\sin }^2}A} \right)^2} + {\left( {{{\cos }^2}A} \right)^2} + 2.{\sin ^2}A.{\cos ^2}A = 1\]
Step – 4 :
On solving and rearranging the terms, we get;
\[{\sin ^4}A + {\cos ^4}A + 2{\sin ^2}A.{\cos ^2}A = 1\]
Step – 5 :
Subtracting \[2{\sin ^2}A.{\cos ^2}A\] from both the sides;
\[{\sin ^4}A + {\cos ^4}A + 2{\sin ^2}A.{\cos ^2}A - 2{\sin ^2}A.{\cos ^2}A\]
\[ = 1 - 2{\sin ^2}A.{\cos ^2}A\]
 i.e. \[{\sin ^4}A + {\cos ^4}A = 1 - 2{\sin ^2}A.{\cos ^2}A\]
Hence L.H.S. = R.H.S.

Note:
$\sin \theta $ and $\cos \theta $ are the main and the very important Trigonometric functions in Mathematics.
With the help of sine and cosine functions i.e. just by knowing the values of $\sin \theta $ and $\cos \theta $ , we are able to find the values of the rest of the Trigonometric functions.
Which are;
\[Tangent \;\theta \], \[\operatorname{Cos} ceant\theta \], \[\operatorname{Sec} ant\;\theta \], \[Cotangent\;\theta \], given that the \[\;\theta \] i.e. the angle must be the same in all these Trigonometric functions as
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ ; $\cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{{\cos \theta }}{{\sin \theta }}$
$\cos ec \theta = \dfrac{1}{{\sin \theta }}$ ; $\cos \theta = \dfrac{1}{{\cos \theta }}$
In this problem,
In step 1, we used the property as the angle is same i.e. A. So,
${\sin ^2}A + {\cos ^2}A = 1$
In step 3, we used the identity
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Hence, in step 5; after rearranging the terms and solving everything, we get
\[{\sin ^4}A + {\cos ^4}A = 1 - 2{\cos ^2}A.{\sin ^2}A\]