## Trigonometry (11th Edition) Clone

$$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$ The proof of identity verification is shown below.
$$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$ From the left side: $$X=\cot\theta-\tan\theta$$ $$X=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$ (according to quotient identities) $$X=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$ Now $\sin^2\theta=1-\cos^2\theta$ (Pythagorean identity) can be replaced into $X$ $$X=\frac{\cos^2\theta-(1-\cos^2\theta)}{\sin\theta\cos\theta}$$ $$X=\frac{\cos^2\theta-1+\cos^2\theta}{\sin\theta\cos\theta}$$ $$X=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$ Therefore, $$\cot\theta-\tan\theta=\frac{2\cos^2\theta-1}{\sin\theta\cos\theta}$$ The equation is an identity as a result.