Verify that the equation given below is an identity. (Hint: $\cos 2 x=\cos (x+x)$.)
\[
\cos 2 x=\frac{\cot ^{2} x-1}{\cot ^{2} x+1}
\]
Rewrite the the $\cot ^{2} x$ terms on the right side of the equation to be in terms of sine and cosine.
\[
\cos 2 x=\frac{\square-1}{\square+1}
\]

Step 1 ：We are given the equation \(\cos 2 x=\frac{\cot ^{2} x-1}{\cot ^{2} x+1}\) and asked to verify that it is an identity.

Step 2 ：We know that \(\cos 2 x=\cos (x+x)\).

Step 3 ：We can rewrite the \(\cot ^{2} x\) terms on the right side of the equation to be in terms of sine and cosine. This gives us \(\cos 2 x=\frac{\square-1}{\square+1}\).

Step 4 ：The cotangent function is the reciprocal of the tangent function, which is the ratio of sine to cosine. Therefore, \(\cot ^{2} x\) can be rewritten as \(\frac{\cos ^{2} x}{\sin ^{2} x}\). Substituting this into the equation, we get \(\cos 2 x=\frac{\frac{\cos ^{2} x}{\sin ^{2} x}-1}{\frac{\cos ^{2} x}{\sin ^{2} x}+1}\).

Step 5 ：By simplifying the equation, we can verify that the two expressions are indeed equal, which means the original equation is an identity.

Step 6 ：\(\boxed{\text{Final Answer: The equation } \cos 2 x=\frac{\cot ^{2} x-1}{\cot ^{2} x+1} \text{ is indeed an identity.}}\)