## Precalculus (6th Edition) Blitzer

$\frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}$ By expressing the terms in the numerator and denominator with the least common denominator $\sin x\cos x$, we get: \begin{align} & \frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}=\frac{\cos x\left( \sin x+\cos x \right)-\sin x\left( \cos x-\sin x \right)}{\sin x\cos x} \\ & =\frac{\cos x\sin x+{{\cos }^{2}}x-\sin x\cos x+{{\sin }^{2}}x}{\sin x\cos x} \\ & =\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x\cos x} \end{align} Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}x+{{\cos }^{2}}x=1$; then, the above expression can be further simplified as: $\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\sin x\cos x}=\frac{1}{\sin x\cos x}$ Then, rewrite the above expression as: $\frac{1}{\sin x\cos x}=\frac{1}{\sin x}.\frac{1}{\cos x}$ By using the reciprocal identity of trigonometry $\csc x=\frac{1}{\sin x}$ and $\sec x=\frac{1}{\cos x}$ , and now the above expression can be further simplified as: \begin{align} & \frac{1}{\sin x}.\frac{1}{\cos x}=\csc x.\sec x \\ & =\csc x\sec x \\ & =\sec x\csc x \end{align} Thus, the left side of the expression is equal to the right side, which is $\frac{\sin x+\cos x}{\sin x}-\frac{\cos x-\sin x}{\cos x}=\sec x\csc x$.