What is the value of ${\cos ^2}A{\left( {3 - 4{{\cos }^2}A} \right)^2} + {\sin ^2}A{\left( {3 - 4{{\sin }^2}A} \right)^2}$?
A. $\cos 4A$
B. $\sin 4A$
C. $1$
D. None of these

Hint: Use the formula ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ for the terms which are in brackets. Then multiply and simplify using the identity ${\cos ^2}A + {\sin ^2}A = 1$.

Formula Used:
${\cos ^2}A + {\sin ^2}A = 1$
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
${a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)$

Complete step by step solution:
The given expression is ${\cos ^2}A{\left( {3 - 4{{\cos }^2}A} \right)^2} + {\sin ^2}A{\left( {3 - 4{{\sin }^2}A} \right)^2}$
Use the formula ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
Substituting $a = 3$ and $b = 4{\cos ^2}A$, we get
${\left( {3 - 4{{\cos }^2}A} \right)^2} = {\left( 3 \right)^2} - 2 \times 3 \times 4{\cos ^2}A + {\left( {4{{\cos }^2}A} \right)^2}$
$ = 9 - 24{\cos ^2}A + 16{\cos ^4}A$
Substituting $a = 3$ and $b = 4{\sin ^2}A$, we get
${\left( {3 - 4{{\sin }^2}A} \right)^2} = {\left( 3 \right)^2} - 2 \times 3 \times 4{\sin ^2}A + {\left( {4{{\sin }^2}A} \right)^2}$
$ = 9 - 24{\sin ^2}A + 16{\sin ^4}A$
Substituting these in the given expression, we get
${\cos ^2}A\left( {9 - 24{{\cos }^2}A + 16{{\cos }^4}A} \right) + {\sin ^2}A{\left( {9 - 24{{\sin }^2}A + 16{{\sin }^4}A} \right)^2}$
Use the formula ${a^m} \cdot {a^n} = {a^{m + n}}$
$ = 9{\cos ^2}A - 24{\cos ^4}A + 16{\cos ^6}A + 9{\sin ^2}A - 24{\sin ^4}A + 16{\sin ^6}A$
Arrange the terms.
$ = 9{\cos ^2}A + 9{\sin ^2}A - 24{\cos ^4}A - 24{\sin ^4}A + 16{\cos ^6}A + 16{\sin ^6}A$
Take common.
$ = 9\left( {{{\cos }^2}A + {{\sin }^2}A} \right) - 24\left( {{{\cos }^4}A + {{\sin }^4}A} \right) + 16\left( {{{\cos }^6}A + {{\sin }^6}A} \right)$
Simplify the terms inside the brackets.
For the terms inside the first bracket, use the identity ${\cos ^2}A + {\sin ^2}A = 1$
For the terms inside the second bracket, ${\cos ^4}A + {\sin ^4}A = {\left( {{{\cos }^2}A} \right)^2} + {\left( {{{\sin }^2}A} \right)^2}$
Use the identity ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$
Substituting $a = {\cos ^2}A$ and $b = {\sin ^2}A$, we get
${\cos ^4}A + {\sin ^4}A = {\left( {{{\cos }^2}A + {{\sin }^2}A} \right)^2} - 2{\cos ^2}A{\sin ^2}A$
Use the identity ${\cos ^2}A + {\sin ^2}A = 1$
$ \Rightarrow {\cos ^4}A + {\sin ^4}A = 1 - 2{\cos ^2}A{\sin ^2}A$
For the terms inside the third bracket, ${\cos ^6}A + {\sin ^6}A = {\left( {{{\cos }^2}A} \right)^3} + {\left( {{{\sin }^2}A} \right)^3}$
Use the identity ${a^3} + {b^3} = {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right)$
Substituting $a = {\cos ^2}A$ and $b = {\sin ^2}A$, we get
${\cos ^6}A + {\sin ^6}A = {\left( {{{\cos }^2}A + {{\sin }^2}A} \right)^3} - 3{\cos ^2}A{\sin ^2}A\left( {{{\cos }^2}A + {{\sin }^2}A} \right)$
Use the identity ${\cos ^2}A + {\sin ^2}A = 1$
$ \Rightarrow {\cos ^6}A + {\sin ^6}A = 1 - 3{\cos ^2}A{\sin ^2}A$
Substituting these expressions, we get
The given expression $ = 9\left( 1 \right) - 24\left( {1 - 2{{\cos }^2}A{{\sin }^2}A} \right) + 16\left( {1 - 3{{\cos }^2}A{{\sin }^2}A} \right)$
$ = 9 - 24 + 48{\cos ^2}A{\sin ^2}A + 16 - 48{\cos ^2}A{\sin ^2}A$
Cancel the term $48{\cos ^2}A{\sin ^2}A$
$ = 9 - 24 + 16 = 1$

Option ‘C’ is correct

Note: Alternate method for finding the value of the given expression. Make both the terms as a perfect square expression. Then use trigonometric identities.
${\cos ^2}A{\left( {3 - 4{{\cos }^2}A} \right)^2} + {\sin ^2}A{\left( {3 - 4{{\sin }^2}A} \right)^2}$
$ = {\left\{ {\cos A\left( {3 - 4{{\cos }^2}A} \right)} \right\}^2} + {\left\{ {\sin A\left( {3 - 4{{\sin }^2}A} \right)} \right\}^2}$
$ = {\left( {3\cos A - 4{{\cos }^3}A} \right)^2} + {\left( {3\sin A - 4{{\sin }^3}A} \right)^2}$
Use the identities $4{\cos ^3}A - 3\cos A = \cos 3A$ and $3\sin A - 4{\sin ^3}A = \sin 3A$
$ = {\left( {\cos 3A} \right)^2} + {\left( {\sin 3A} \right)^2}$
$ = {\cos ^2}3A + {\sin ^2}3A$
Use the identity ${\cos ^2}A + {\sin ^2}A = 1$