Question

# Proof trigonometric identities.\sec(x)+\tan(x)=\frac{\cos(x)}{1-\sin(x)}

Trigonometric equation and identitie
Proof trigonometric identities.
$$\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}={\frac{{{\cos{{\left({x}\right)}}}}}{{{1}-{\sin{{\left({x}\right)}}}}}}$$

2021-09-13
The given equation is $$\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}={\frac{{{\cos{{\left({x}\right)}}}}}{{{1}-{\sin{{\left({x}\right)}}}}}}$$
Let choose $$\displaystyle{\sec{{\left({x}\right)}}}+{\tan{{\left({x}\right)}}}$$ to prove $$\displaystyle{\frac{{{\cos{{x}}}}}{{{1}-{\sin{{x}}}}}}$$ to prove the identity.
$$\displaystyle{\sec{{x}}}+{\tan{{x}}}={\frac{{{1}}}{{{\cos{{x}}}}}}+{\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}$$
($$\displaystyle\because{\sec{{x}}}={\frac{{{1}}}{{{\cos{{x}}}}}},{\tan{{x}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}$$)
$$\displaystyle={\frac{{{1}+{\sin{{x}}}}}{{{\cos{{x}}}}}}$$
$$\displaystyle={\frac{{{1}+{\sin{{x}}}}}{{{\cos{{x}}}}}}\cdot{\frac{{{\cos{{x}}}}}{{{\cos{{x}}}}}}$$
$$\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{{\cos}^{{2}}{x}}}}}$$
On further simplification,
$$\displaystyle{\sec{{x}}}+{\tan{{x}}}={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{\left({1}-{{\sin}^{{2}}{x}}\right)}}}}$$
$$\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{1}^{{2}}-{{\sin}^{{2}}{x}}}}}$$
$$\displaystyle={\frac{{{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}}}{{{\left({1}+{\sin{{x}}}\right)}{\left({1}-{\sin{{x}}}\right)}}}}$$
$$\displaystyle={\frac{{{\cos{{x}}}}}{{{\left({1}-{\sin{{x}}}\right)}}}}$$
Hence, the given identity is proved.