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

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

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Marcus Herman
Use
$$\displaystyle{\frac{{{\sec{{\frac{{x}}{{2}}}}}}}{{{\tan{{\frac{{x}}{{2}}}}}+{1}}}}\equiv{\frac{{{1}}}{{{\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}}}}$$
$$\displaystyle{\left({\sin{{\frac{{x}}{{2}}}}}+{\cos{{\frac{{x}}{{2}}}}}\right)}^{{2}}=?$$
###### Not exactly what youâ€™re looking for?
stomachdm
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.
Vasquez

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