Prove the identity $\frac{\cos 2 x}{\sin ^{2} x+\frac{1}{2} \sin 2 x}=\cot x-1$

Step 1 ：First, we need to understand the problem. We are asked to prove the identity $\frac{\cos 2 x}{\sin ^{2} x+\frac{1}{2} \sin 2 x}=\cot x-1$. This means we need to show that the left-hand side (LHS) of the equation is equal to the right-hand side (RHS).

Step 2 ：Let's start with the LHS of the equation, $\frac{\cos 2 x}{\sin ^{2} x+\frac{1}{2} \sin 2 x}$. We can simplify this expression by using the double angle formula for cosine, which is $\cos 2x = \cos^2 x - \sin^2 x$.

Step 3 ：Substitute $\cos 2x$ with $\cos^2 x - \sin^2 x$ in the LHS, we get $\frac{\cos^2 x - \sin^2 x}{\sin ^{2} x+\frac{1}{2} \sin 2 x}$.

Step 4 ：Next, we can simplify the denominator by using the double angle formula for sine, which is $\sin 2x = 2\sin x\cos x$. Substitute $\sin 2x$ with $2\sin x\cos x$ in the denominator, we get $\frac{\cos^2 x - \sin^2 x}{\sin ^{2} x+\sin x\cos x}$.

Step 5 ：We can further simplify the expression by dividing both the numerator and the denominator by $\sin^2 x$. We get $\frac{\cot^2 x - 1}{1+\cot x}$.

Step 6 ：The expression $\frac{\cot^2 x - 1}{1+\cot x}$ can be rewritten as $\cot x - 1$ by using the method of partial fraction decomposition.

Step 7 ：Therefore, we have shown that the LHS of the equation is equal to the RHS, which is $\cot x - 1$. Hence, the identity $\frac{\cos 2 x}{\sin ^{2} x+\frac{1}{2} \sin 2 x}=\cot x-1$ is proven.