Answer
$$\frac{-1}{\tan\alpha-\sec\alpha}+\frac{-1}{\tan\alpha+\sec\alpha}=2\tan\alpha$$
The left side is equal to the right side. This is an identity.
Work Step by Step
$$\frac{-1}{\tan\alpha-\sec\alpha}+\frac{-1}{\tan\alpha+\sec\alpha}=2\tan\alpha$$
We would simplify the left side.
$$A=\frac{-1}{\tan\alpha-\sec\alpha}+\frac{-1}{\tan\alpha+\sec\alpha}$$
$$A=\frac{-(\tan\alpha+\sec\alpha)-(\tan\alpha-\sec\alpha)}{(\tan\alpha-\sec\alpha)(\tan\alpha+\sec\alpha)}$$
$$A=\frac{-\tan\alpha-\sec\alpha-\tan\alpha+\sec\alpha}{\tan^2\alpha-\sec^2\alpha}$$ (since $(a-b)(a+b)=a^2-b^2$)
$$A=\frac{-2\tan\alpha}{\tan^2\alpha-\sec^2\alpha}$$
Now we transform $\tan\alpha$ and $\sec\alpha$ according to the following identities:
$$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\hspace{2cm}\sec\alpha=\frac{1}{\cos\alpha}$$
Therefore, $$\tan^2\alpha-\sec^2\alpha=\frac{\sin^2\alpha}{\cos^2\alpha}-\frac{1}{\cos^2\alpha}=\frac{\sin^2\alpha-1}{\cos^2\alpha}=\frac{-(1-\sin^2\alpha)}{\cos^2\alpha}=\frac{-\cos^2\alpha}{\cos^2\alpha}=-1$$ (since $1-\sin^2\alpha=\cos^2\alpha$, according to a Pythagorean identity)
Hence, $A$ would be
$$A=\frac{-2\tan\alpha}{-1}$$
$$A=2\tan\alpha$$
We thus have proved that the expression is an identity by making the left side equal to the right side.