## Trigonometry (11th Edition) Clone

$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ The equation is an identity. The proof is below.
$$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ We solve from the left side, as it is more complex. $$X=\frac{2\cos2\theta}{\sin2\theta}$$ - We replace $\cos2\theta$ and $\sin2\theta$ with the following identities from Double-Angle identities: $$\cos2\theta=\cos^2\theta-\sin^2\theta$$ $$\sin2\theta=2\sin\theta\cos\theta$$ Therefore, $$X=\frac{2(\cos^2\theta-\sin^2\theta)}{2\sin\theta\cos\theta}$$ $$X=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}$$ $$X=\frac{\cos^2\theta}{\sin\theta\cos\theta}-\frac{\sin^2\theta}{\sin\theta\cos\theta}$$ $$X=\frac{\cos\theta}{\sin\theta}-\frac{\sin\theta}{\cos\theta}$$ - From Quotient Identities: $\frac{\cos\theta}{\sin\theta}=\cot\theta$ and $\frac{\sin\theta}{\cos\theta}=\tan\theta$. Therefore, $$X=\cot\theta-\tan\theta$$ So, $$\frac{2\cos2\theta}{\sin2\theta}=\cot\theta-\tan\theta$$ We conclude that the equation is an identity.