Trigonometry (11th Edition) Clone

$$\frac{\cos(\alpha-\beta)}{\cos\alpha\sin\beta}=\tan\alpha+\cot\beta$$ The equation is an identity.
$$\frac{\cos(\alpha-\beta)}{\cos\alpha\sin\beta}=\tan\alpha+\cot\beta$$ The left side seems more complicated, so we would deal with it first. $$\frac{\cos(\alpha-\beta)}{\cos\alpha\sin\beta}$$ (apply the identity of cosine of a difference) $$=\frac{\cos\alpha\cos\beta+\sin\alpha\sin\beta}{\cos\alpha\sin\beta}$$ $$=\frac{\cos\alpha\cos\beta}{\cos\alpha\sin\beta}+\frac{\sin\alpha\sin\beta}{\cos\alpha\sin\beta}$$ $$=\frac{\cos\beta}{\sin\beta}+\frac{\sin\alpha}{\cos\alpha}$$ $$=\cot\beta+\tan\alpha$$ Therefore, both sides are equal. The equation has been verified to be an identity.