How do you simplify $ 1 - 4{\sin ^2}x{\cos ^2}x $ ?

Hint: In this question we need to simplify $ 1 - 4{\sin ^2}x{\cos ^2}x $ . In order to simplify $ 1 - 4{\sin ^2}x{\cos ^2}x $ , we will use trigonometric identities such as $ \sin 2x = 2\sin x\cos x $ and $ {\sin ^2}x + {\cos ^2}x = 1 $ . By using these identities and evaluating it, we will determine the required answer.

Complete step-by-step answer:
Here we need to simplify $ 1 - 4{\sin ^2}x{\cos ^2}x $ .
The given term is $ 1 - 4{\sin ^2}x{\cos ^2}x $ .
 $ = 1 - {\left( {2\sin x\cos x} \right)^2} $
Now, we know that $ \sin 2x = 2\sin x\cos x $ .
Thus, by substituting the value, we have,
 $ = 1 - {\left( {\sin 2x} \right)^2} $
 $ = 1 - \left( {{{\sin }^2}2x} \right) $
Again from trigonometric identities, we have,
 $ {\sin ^2}x + {\cos ^2}x = 1 $
  $ {\cos ^2}x = 1 - {\sin ^2}x $
Therefore, by substituting, we have,
 $ = \left( {{{\cos }^2}2x} \right) $
Hence, by simplifying $ 1 - 4{\sin ^2}x{\cos ^2}x $ we get $ {\cos ^2}2x $ .
So, the correct answer is “ $ {\cos ^2}2x $ ”.

Note: In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles.
Sine, cosine, secant, and cosecant have period $2\pi$ while tangent and cotangent have period $\pi$. Identities for negative angles. Sine, tangent, cotangent, and cosecant are odd functions while cosine and secant are even functions