Prove that \frac{\sin a - \cos a + 1}{\sin a} = 2 \cos \frac{a}{2}

To prove the given equation, we can start by manipulating the left-hand side (LHS) of the equation. We have:

LHS = \frac{\sin a - \cos a + 1}{\sin a}

We can rewrite the numerator using trigonometric identities. Recall the following:

\sin a = 2 \sin \frac{a}{2} \cos \frac{a}{2}
\cos a = \cos^2 \frac{a}{2} - \sin^2 \frac{a}{2}
1 = \cos^2 \frac{a}{2} + \sin^2 \frac{a}{2}

Substituting these identities into the LHS:

LHS = \frac{2 \sin \frac{a}{2} \cos \frac{a}{2} - (\cos^2 \frac{a}{2} - \sin^2 \frac{a}{2}) + (\cos^2 \frac{a}{2} + \sin^2 \frac{a}{2})}{2 \sin \frac{a}{2} \cos \frac{a}{2}}

Simplifying the numerator:

LHS = \frac{2 \sin \frac{a}{2} \cos \frac{a}{2} + 2 \sin^2 \frac{a}{2}}{2 \sin \frac{a}{2} \cos \frac{a}{2}}

Factoring out (2 \sin \frac{a}{2}) from the numerator:

LHS = \frac{2 \sin \frac{a}{2} (\cos \frac{a}{2} + \sin \frac{a}{2})}{2 \sin \frac{a}{2} \cos \frac{a}{2}}

Assuming (\sin \frac{a}{2} \neq 0), we can cancel the common factor:

LHS = \frac{\cos \frac{a}{2} + \sin \frac{a}{2}}{\cos \frac{a}{2}}

Now, divide each term in the numerator by (\cos \frac{a}{2}):

LHS = 1 + \frac{\sin \frac{a}{2}}{\cos \frac{a}{2}} = 1 + \tan \frac{a}{2}

This does not directly match the given solution. There might be a mistake in the original question or a missing condition. Let's try another approach. The provided solution suggests that the original question might be incorrect. There may be a typo or a different intended equation.