# Verify the identity secx-secx sin^2x= cosx

Question
Verify the identity $$\displaystyle{\sec{{x}}}-{\sec{{x}}}{{\sin}^{{2}}{x}}={\cos{{x}}}$$

2021-02-12
$$\displaystyle{\sec{{x}}}-{\sec{{x}}}{{\sin}^{{2}}{x}}={\cos{{x}}}$$
$$\displaystyle{\sec{{x}}}{\left({1}-{{\sin}^{{2}}{x}}\right)}={\cos{{x}}}$$
because $$\displaystyle{{\sin}^{{x}}+}{{\cos}^{{x}}=}{1},{1}-{{\sin}^{{2}}{x}}={{\cos}^{{2}}{x}}$$
$$\displaystyle{\sec{{x}}}{{\cos}^{{2}}{x}}={\cos{{x}}}$$
$$\displaystyle{\left(\frac{{I}}{{\cos{{x}}}}\right)}{{\cos}^{{2}}{x}}={\cos{{x}}}$$
$$\displaystyle{\cos{{x}}}={\cos{{x}}}$$

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