## Precalculus (6th Edition) Blitzer

We have the expression on the left side $\frac{1}{\sin t-1}+\frac{1}{\sin t+1}$, which can be simplified by multiplying and dividing the expression by $\sin t+1$ and $\sin t-1$, and then further simplifying by using the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ \begin{align} & \frac{1}{\sin t-1}+\frac{1}{\sin t+1}=\frac{1}{\sin t-1}.\frac{\sin t+1}{\sin t+1}+\frac{1}{\sin t+1}.\frac{\sin t-1}{\sin t-1} \\ & =\frac{\sin t+1}{{{\sin }^{2}}t-1}+\frac{\sin t-1}{{{\sin }^{2}}t-1} \\ & =\frac{\sin t+1+\sin t-1}{{{\sin }^{2}}t-1} \\ & =\frac{2\sin t}{{{\sin }^{2}}t-1} \end{align} We know the Pythagorean identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$; therefore, ${{\sin }^{2}}x-1=-{{\cos }^{2}}x$ and using the quotient identity $\tan x=\frac{\sin x}{\cos x}$ and the reciprocal identities $\frac{1}{\cos x}=\sec x$, the expression can be simplified as: \begin{align} & \frac{2\sin t}{{{\sin }^{2}}t-1}=\frac{2\sin t}{-{{\cos }^{2}}t} \\ & =-2.\frac{\sin t}{\cos t}.\frac{1}{\cos t} \\ & =-2\tan t.\sec t \end{align} Hence, the expression on the left side is equal to the expression on the right side.