How to prove $\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos(A)^2}$ I am unable to prove this trigonometric identity
$$\frac{\tan (A)}{\tan (A)}+\frac{\cot (A)}{\cot (A)}=\frac{1}{1-2\cos^2(A)}$$
I have tried to transform the left-hand side and stuck with this
$$\frac{2\sin(A)\cos(A)}{\sin(A)\cos(A)}$$
And I have tried to transform the right-hand side by changing the $$2\cos^2(A)$$ to $$\frac{2}{\sec^2(A)}$$, and used the trigonometric identity $$1+\tan^2(A)=\sec^2(A)$$ and got this instead
$$\frac{1+\tan^2(A)}{\tan^2(A)-1}$$ which I can transform to $$\frac{\cot(A)+\tan(A)}{\tan(A)-\cot(A)}$$.
I cannot get both sides equal, help please?

One way we can prove the identity false is as follows:
$$\begin {align}
\dfrac {\tan A} {\tan A} + \dfrac{\cot A}{\cot A} = \dfrac {1}{1-2\cos^2 2A} \\
2 = \dfrac {1}{1-2\cos^2 2A} \\
2 (1-2\cos^2 2A) = 1 \\
2 - 4\cos^2 2A = 1 \\
- 4\cos^2 2A = \dfrac {1}{2} \\
\cos^2 2A = -\dfrac {1}{8}
\end {align}$$
Since the last line would require us to take the square root of a negative number, $A$ does not exist, and the identity is false.
