# How to prove the following: tan^2x+1+tanx secx = 1+sin x/ cos^2x

Trigonometric equation and identitie
How to prove the following:
$$\displaystyle{{\tan}^{{2}}{x}}+{1}+{\tan{{x}}}{\sec{{x}}}={1}+\frac{{\sin{{x}}}}{{{\cos}^{{2}}{x}}}$$

$$\displaystyle{\tan{{x}}}\cdot{\sec{{x}}}=\frac{{\sin{{x}}}}{{{\cos}^{{2}}{x}}}{\quad\text{or}\quad}{\sin{{x}}}\cdot{{\sec}^{{2}}{x}}.{{\tan}^{{2}}{x}}+{1}={{\sec}^{{2}}{x}}.$$
So we have $$\displaystyle{{\sec}^{{2}}{x}}+{{\sec}^{{2}}{x}}{\sin{{x}}}={{\sec}^{{2}}{x}}{\left({1}+{\sin{{x}}}\right)}=\frac{{{1}+{\sin{{x}}}}}{{{\cos}^{{2}}{x}}}.$$
(To prove $$\displaystyle{{\sec}^{{2}}{x}}={{\tan}^{{2}}{x}}+{1}$$, start with $$\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}$$, then divide through by $$\displaystyle{{\cos}^{{2}}{x}}:{{\tan}^{{2}}{x}}+{1}={{\sec}^{{2}}{x}}$$ because secx is $$\displaystyle\frac{{1}}{{\cos{{x}}}}$$.)