## Trigonometry (11th Edition) Clone

$$\tan x+\cot x=2\csc2x$$ The equation is an identity, as the left side is equal to the right one.
$$\tan x+\cot x=2\csc2x$$ The right side would be taken into consideration first. $$X=2\csc2x$$ We know that $\csc\theta=\frac{1}{\sin\theta}$, so $\csc2x=\frac{1}{\sin2x}$. $$X=\frac{2}{\sin2x}$$ Now $\sin 2x=2\sin x\cos x$, as stated in Double-Angle Identities. $$X=\frac{2}{2\sin x\cos x}$$ $$X=\frac{1}{\sin x\cos x}$$ We can rewrite $1$ as $\sin^2x+\cos^2x$, as they also equal $1$. $$X=\frac{\sin^2x+\cos^2x}{\sin x\cos x}$$ $$X=\frac{\sin^2x}{\sin x\cos x}+\frac{\cos^2x}{\sin x\cos x}$$ $$X=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$$ Applying Quotient Identities: $\frac{\sin x}{\cos x}=\tan x$ and $\frac{\cos x}{\sin x}=\cot x$. $$X=\tan x+\cot x$$ Thus, $$\tan x+\cot x=2\csc2x$$ The equation is an identity, as the left side is equal to the right one.