How to simplify 11+sin⁡x+11−sin⁡x?

Question

inventivx64 · Accepted Answer

Initially, multiply the first fraction by \(\displaystyle\text{}{1}-{\sin{{x}}}\text{}\) and the second by \(\displaystyle\text{}{1}+{\sin{{x}}}\text{}\)
\(\displaystyle{E}=\frac{{{1}-{\sin{{x}}}}}{{{\left({1}+{\sin{{x}}}\right)}\cdot{\left({1}-{\sin{{x}}}\right)}}}+\frac{{{1}+{\sin{{x}}}}}{{{\left({1}+{\sin{{x}}}\right)}\cdot{\left({1}-{\sin{{x}}}\right)}}}\)
\(\displaystyle{E}=\frac{{{1}\cancel{{-{\sin{{x}}}}}+{1}\cancel{{+{\sin{{x}}}}}}}{{{\left({1}+{\sin{{x}}}\right)}\cdot{\left({1}-{\sin{{x}}}\right)}}}=\frac{{2}}{{{\left({1}+{\sin{{x}}}\right)}\cdot{\left({1}-{\sin{{x}}}\right)}}}\)
Make use of the algebraic identity \(\displaystyle{a}^{{2}}-{b}^{{2}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\). In your case,
\(\displaystyle{a}={1}\) and
\(\displaystyle{b}={\sin{{x}}}\)
Consequently, the expression used as a denominator will change to
\(\displaystyle{\left({1}+{\sin{{x}}}\right)}\cdot{\left({1}-{\sin{{x}}}\right)}={1}^{{{2}}}-{\left({\sin{{x}}}\right)}^{{{2}}}={1}-{{\sin}^{{2}}{x}}\)
Hence, \(\displaystyle{E}\) will be
\(\displaystyle{E}=\frac{{2}}{{{1}-{{\sin}^{{2}}{x}}}}\)
Remember that \(\displaystyle{1}-{{\sin}^{{2}}{x}}={{\cos}^{{2}}{x}}\), so the final form of \(\displaystyle{E}\) will be
\(\displaystyle{E}={\color{green}{\frac{{2}}{{{\cos}^{{2}}{x}}}}}\)

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