Question

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

Trigonometric equation and identitie
ANSWERED
asked 2021-03-02
How to prove the following:
\(\displaystyle{{\tan}^{{2}}{x}}+{1}+{\tan{{x}}}{\sec{{x}}}={1}+\frac{{\sin{{x}}}}{{{\cos}^{{2}}{x}}}\)

Answers (1)

2021-03-03
\(\displaystyle{\tan{{x}}}\cdot{\sec{{x}}}=\frac{{\sin{{x}}}}{{{\cos}^{{2}}{x}}}{\quad\text{or}\quad}{\sin{{x}}}\cdot{{\sec}^{{2}}{x}}.{{\tan}^{{2}}{x}}+{1}={{\sec}^{{2}}{x}}.\)
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 \(\displaystyle\frac{{1}}{{\cos{{x}}}}\).)
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