## Trigonometry (11th Edition) Clone

$$\frac{-1}{\tan\alpha-\sec\alpha}+\frac{-1}{\tan\alpha+\sec\alpha}=2\tan\alpha$$ The left side is equal to the right side. This is an identity.
$$\frac{-1}{\tan\alpha-\sec\alpha}+\frac{-1}{\tan\alpha+\sec\alpha}=2\tan\alpha$$ 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.