## Trigonometry (11th Edition) Clone

$$\frac{\cot A-\tan A}{\cot A+\tan A}=\cos 2A$$ The equation is an identity. The proof is in the work step by step.
$$\frac{\cot A-\tan A}{\cot A+\tan A}=\cos 2A$$ The left side, for being more complex, has to be tackled first. $$X=\frac{\cot A-\tan A}{\cot A+\tan A}$$ As the right side is $\cos2A$, which would involve $\sin A$ and $\cos A$, probably here we need to rewrite $$\cot A=\frac{\cos A}{\sin A}\hspace{2cm}\tan A=\frac{\sin A}{\cos A}$$ $$X=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}$$ $$X=\frac{\frac{\cos^2A-\sin^2A}{\sin A\cos A}}{\frac{\cos^2A+\sin^2A}{\sin A\cos A}}$$ The $\sin A\cos A$ in the numerator and the $\sin A\cos A$ in the denominator would eliminate each other. $$X=\frac{\cos^2A-\sin^2A}{\cos^2A+\sin^2A}$$ - From Double-Angle Identity for cosine: $\cos^2A-\sin^2A=\cos2A$ - From Pythagorean Identity: $\cos^2A+\sin^2A=1$ Therefore, $$X=\frac{\cos2A}{1}$$ $$X=\cos2A$$ So, $$\frac{\cot A-\tan A}{\cot A+\tan A}=\cos 2A$$ The equation is an identity.