Trigonometry (11th Edition) Clone

$$(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)=\cot\alpha-\tan\alpha$$ The expression is proved below to be an identity.
$$(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)=\cot\alpha-\tan\alpha$$ We examine the left side. $$A=(\sec\alpha+\csc\alpha)(\cos\alpha-\sin\alpha)$$ $\sec\alpha$ and $\csc\alpha$ can be rewritten as follows. $$\sec\alpha=\frac{1}{\cos\alpha}\hspace{2cm}\csc\alpha=\frac{1}{\sin\alpha}$$ which means, $$A=\Big(\frac{1}{\cos\alpha}+\frac{1}{\sin\alpha}\Big)(\cos\alpha-\sin\alpha)$$ $$A=\frac{\sin\alpha+\cos\alpha}{\sin\alpha\cos\alpha}(\cos\alpha-\sin\alpha)$$ $$A=\frac{(\cos\alpha+\sin\alpha)(\cos\alpha-\sin\alpha)}{\sin\alpha\cos\alpha}$$ $$A=\frac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}$$ (as $(a+b)(a-b)=a^2-b^2$) $$A=\frac{\cos^2\alpha}{\sin\alpha\cos\alpha}-\frac{\sin^2\alpha}{\sin\alpha\cos\alpha}$$ $$A=\frac{\cos\alpha}{\sin\alpha}-\frac{\sin\alpha}{\cos\alpha}$$ $$A=\cot\alpha-\tan\alpha\hspace{1cm}\text{(Quotient Identities)}$$ 2 sides are equal, so the expression is an identity.