## Precalculus (6th Edition) Blitzer

And the given expression on the left side $1-\frac{{{\sin }^{2}}x}{1+\cos x}$ can be simplified by using the Pythagorean identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Thus, ${{\sin }^{2}}x=1-{{\cos }^{2}}x$. Then, further simplify the expression by applying the algebraic formula $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ \begin{align} & 1-\frac{{{\sin }^{2}}x}{1+\cos x}=1-\frac{1-{{\cos }^{2}}x}{1+\cos x} \\ & =1-\frac{\left( 1+\cos x \right)\left( 1-\cos x \right)}{1+\cos x} \\ & =1-\left( 1-\cos x \right) \\ & =\cos x \end{align} Hence, the expression on the left side is equal to the expression on the right side.