How do you prove $ \cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A $ ?

Hint:
The above-given question is of trigonometric identities. So, we will use a trigonometric formula like cos(A + B) = cos A cos B - sin A sin B, and then we will put B = A in the equation to get 2A on the left side of the identity, and then we will get the required result.

Complete step by step answer:
We can see that the above-given question is of trigonometric identity and so we will use trigonometric formulas to prove the above result $ \cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A $ .
We have to prove that $ \cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A $ and from trigonometric formula of cos(A + B) we know that:
 $ \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $
So, when we put B = A in the above equation we will get:
 $ \cos \left( A+A \right)=\cos A\cos A-\sin A\sin A $
After we add the angle and express the terms on the RHS as squares, we will get the final result as
 $ \Rightarrow \cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A $
Hence proved.
Hence, this is our required proof and solution to the above question.

Note:
Students are required to note that we can also solve the above question by making LHS (Left Hand Side) i.e. cos 2A equal to RHS (Right Hand Side) i.e. $ {{\cos }^{2}}A-{{\sin }^{2}}A $.
Since, LHS = cos 2A and RHS = $ {{\cos }^{2}}A-{{\sin }^{2}}A $
We can also write it as:
LHS = $ \cos 2A=\cos \left( A+A \right) $
Now, we know that $ \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B $ :
So, we can write $ \cos 2A=\cos \left( A+A \right)=\cos A\cos A-\sin A\sin A $
 $ \Rightarrow \cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A $ = RHS
Hence, LHS = RHS.
Hence, proved.