Prove the trigonometric identity: $tan(\frac{\theta}{2}) + cot(\frac{\theta}{2}) = 2csc(\theta)$.

1. Problem Description

Prove the trigonometric identity: $tan(\frac{\theta}{2}) + cot(\frac{\theta}{2}) = 2csc(\theta)$.

2. Solution Steps

We will start with the left-hand side of the equation and try to simplify it to the right-hand side.

$tan(\frac{\theta}{2}) + cot(\frac{\theta}{2})$

Since $cot(x) = \frac{1}{tan(x)}$, we can rewrite the expression as:

$tan(\frac{\theta}{2}) + \frac{1}{tan(\frac{\theta}{2})}$

Now, let's find a common denominator:

$\frac{tan^2(\frac{\theta}{2}) + 1}{tan(\frac{\theta}{2})}$

We know that $1 + tan^2(x) = sec^2(x)$. Therefore,

$\frac{sec^2(\frac{\theta}{2})}{tan(\frac{\theta}{2})}$

Now, rewrite in terms of sine and cosine:

$\frac{\frac{1}{cos^2(\frac{\theta}{2})}}{\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})}} = \frac{1}{cos^2(\frac{\theta}{2})} \cdot \frac{cos(\frac{\theta}{2})}{sin(\frac{\theta}{2})} = \frac{1}{cos(\frac{\theta}{2})sin(\frac{\theta}{2})}$

Multiply numerator and denominator by 2:

$\frac{2}{2sin(\frac{\theta}{2})cos(\frac{\theta}{2})}$

Using the double angle formula $sin(2x) = 2sin(x)cos(x)$, we have:

$sin(\theta) = 2sin(\frac{\theta}{2})cos(\frac{\theta}{2})$

Therefore,

$\frac{2}{sin(\theta)}$

Since $csc(\theta) = \frac{1}{sin(\theta)}$,

$2csc(\theta)$

Thus, $tan(\frac{\theta}{2}) + cot(\frac{\theta}{2}) = 2csc(\theta)$.

3. Final Answer

$tan(\frac{\theta}{2}) + cot(\frac{\theta}{2}) = 2csc(\theta)$ is proven.