How can I prove the identity \frac{1}{1+\sin(x)} \equiv \frac{\sec^2 (\frac{x}{2})}{(\tan(\frac{x}{2}) +1)^2}

Pamela Meyer 2022-01-03 Answered
How can I prove the identity
\(\displaystyle{\frac{{{1}}}{{{1}+{\sin{{\left({x}\right)}}}}}}\equiv{\frac{{{{\sec}^{{2}}{\left({\frac{{{x}}}{{{2}}}}\right)}}}}{{{\left({\tan{{\left({\frac{{{x}}}{{{2}}}}\right)}}}+{1}\right)}^{{2}}}}}\)

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Expert Answer

Marcus Herman
Answered 2022-01-04 Author has 2086 answers
Use
\(\displaystyle{\frac{{{\sec{{\frac{{x}}{{2}}}}}}}{{{\tan{{\frac{{x}}{{2}}}}}+{1}}}}\equiv{\frac{{{1}}}{{{\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}}}}\)
Now you have already identified
\(\displaystyle{\left({\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}\right)}^{{2}}=?\)
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stomachdm
Answered 2022-01-05 Author has 2733 answers
We can express any value trigonometric ratio of a given angle \(\displaystyle{2}\theta\) in terms of the tangent of half that angle. For example:
\(\displaystyle{\sin{{2}}}\theta\equiv{\frac{{{2}{\tan{\theta}}}}{{{1}+{{\tan}^{{2}}\theta}}}}\)
\(\displaystyle{\cos{{2}}}\theta\equiv{\frac{{{1}-{{\tan}^{{2}}\theta}}}{{{1}+{{\tan}^{{2}}\theta}}}}\)
\(\displaystyle{\tan{{2}}}\theta\equiv{\frac{{{2}{\tan{\theta}}}}{{{1}-{{\tan}^{{2}}\theta}}}}\)
As an aside, this is useful in integration. So, you may want to use the first identity given to prove your identity.
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Vasquez
Answered 2022-01-09 Author has 9499 answers

Alternately, use half angle theorems:
\(\sin x \equiv \frac{2 \tan(\frac x2)}{1+\tan^2(\frac x2)}\)

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