Certainly! Let's prove the given trigonometric identity step by step:
[tex]\[
\tan \frac{x}{2} + \cot \frac{x}{2} = 2 \csc x
\][/tex]
1. Express [tex]\(\tan\frac{x}{2}\)[/tex] and [tex]\(\cot\frac{x}{2}\)[/tex] using sine and cosine:
[tex]\[
\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}
\][/tex]
[tex]\[
\cot\frac{x}{2} = \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}
\][/tex]
2. Combine [tex]\(\tan\frac{x}{2}\)[/tex] and [tex]\(\cot\frac{x}{2}\)[/tex] over a common denominator:
[tex]\[
\tan\frac{x}{2} + \cot\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} + \frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}
\][/tex]
[tex]\[
= \frac{\sin^2\frac{x}{2} + \cos^2\frac{x}{2}}{\sin\frac{x}{2} \cos\frac{x}{2}}
\][/tex]
3. Use the Pythagorean identity:
[tex]\[
\sin^2\frac{x}{2} + \cos^2\frac{x}{2} = 1
\][/tex]
Therefore:
[tex]\[
\tan\frac{x}{2} + \cot\frac{x}{2} = \frac{1}{\sin\frac{x}{2} \cos\frac{x}{2}}
\][/tex]
4. Recall the double-angle identity for sine:
[tex]\[
\sin x = 2 \sin\frac{x}{2} \cos\frac{x}{2}
\][/tex]
This means:
[tex]\[
\sin\frac{x}{2} \cos\frac{x}{2} = \frac{\sin x}{2}
\][/tex]
5. Substitute this back into the expression:
[tex]\[
\tan\frac{x}{2} + \cot\frac{x}{2} = \frac{1}{\frac{\sin x}{2}} = \frac{2}{\sin x}
\][/tex]
6. Use the definition of cosecant:
[tex]\[
\csc x = \frac{1}{\sin x}
\][/tex]
So:
[tex]\[
\frac{2}{\sin x} = 2 \csc x
\][/tex]
Therefore:
[tex]\[
\tan\frac{x}{2} + \cot\frac{x}{2} = 2 \csc x
\][/tex]
This completes our proof! The identity [tex]\(\tan\frac{x}{2} + \cot\frac{x}{2} = 2 \csc x\)[/tex] has been successfully proven.
