## Trigonometry (11th Edition) Clone

$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ After proving 2 sides are equal to each other, we can state that the equation is an identity.
$$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ From the left first: $$X=\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}$$ - A recount of all the identities here: $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$ $$\cos(y-x)=\cos y\cos x+\sin y\sin x=\cos x\cos y+\sin x\sin y$$ $$\sin(x+y)=\sin x\cos y+\cos x\sin y$$ $$\sin(y-x)=\sin y\cos x-\cos y\sin x=\cos x\sin y-\sin x\cos y$$ Therefore: - Numerator: $$\cos(x+y)+\cos(y-x)=\cos x\cos y-\sin x\sin y+\cos x\cos y+\sin x\sin y$$ $$\cos(x+y)+\cos(y-x)=2\cos x\cos y$$ - Denominator: $$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-(\cos x\sin y-\sin x\cos y)$$ $$\sin(x+y)-\sin(y-x)=\sin x\cos y+\cos x\sin y-\cos x\sin y+\sin x\cos y$$ $$\sin(x+y)-\sin(y-x)=2\sin x\cos y$$ Apply back to $X$: $$X=\frac{2\cos x\cos y}{2\sin x\cos y}$$ $$X=\frac{\cos x}{\sin x}$$ $$X=\cot x\hspace{1cm}\text{Quotient Identities}$$ Therefore, $$\frac{\cos(x+y)+\cos(y-x)}{\sin(x+y)-\sin(y-x)}=\cot x$$ That makes the equation eligible to be an identity.