## Trigonometry (11th Edition) Clone

$$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$ The proof is explained in the Work step by step.
$$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$ The left side is more complicated, which is the reason why we need to deal with it first. $$X=\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}$$ - The following identities would be applied to rewrite $\tan\theta$ and $\cot\theta$. $$\tan\theta=\frac{\sin\theta}{\cos\theta}\hspace{2cm}\cot\theta=\frac{\cos\theta}{\sin\theta}$$ $$X=\frac{\frac{\sin\theta}{\cos\theta}-\frac{\cos\theta}{\sin\theta}}{\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}}$$ $$X=\frac{\frac{\sin^2\theta-\cos^2\theta}{\cos\theta\sin\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\cos\theta\sin\theta}}$$ $$X=\frac{\sin^2\theta-\cos^2\theta}{\sin^2\theta+\cos^2\theta}$$ - Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$ $$X=\frac{\sin^2\theta-\cos^2\theta}{1}$$ $$X=\sin^2\theta-\cos^2\theta$$ - Again, Pythagorean identity: $\sin^2\theta=1-\cos^2\theta$ $$X=1-\cos^2\theta-\cos^2\theta$$ $$X=1-2\cos^2\theta$$ That means $$\frac{\tan\theta-\cot\theta}{\tan\theta+\cot\theta}=1-2\cos^2\theta$$ The equation is an identity.