Prove that : `(sin i)/(1+cos i)=(1-cos i)/(sin i)`

Answer

To prove that \(\frac{\sin i}{1 + \cos i} = \frac{1 - \cos i}{\sin i}\), we will start with the left-hand side (LHS) and manipulate it to show that it is equal to the right-hand side (RHS).

### Step-by-Step Solution:

1. **Start with the LHS:**
\[ \text{LHS} = \frac{\sin i}{1 + \cos i} \]

2. **Use the Pythagorean identity:**
We know that \( \sin^2 i + \cos^2 i = 1 \). Therefore, we can express \( \sin^2 i \) as:
\[ \sin^2 i = 1 - \cos^2 i \]

3. **Rewrite the denominator:**
We can rewrite \(1 - \cos^2 i\) as \((1 - \cos i)(1 + \cos i)\) using the difference of squares:
\[ 1 - \cos^2 i = (1 - \cos i)(1 + \cos i) \]

4. **Substitute in the LHS:**
Substitute \(1 - \cos^2 i\) into the LHS:
\[ \text{LHS} = \frac{\sin i}{1 + \cos i} = \frac{\sin i}{\frac{1 - \cos^2 i}{1 - \cos i}} = \frac{\sin i \cdot (1 - \cos i)}{1 - \cos^2 i} \]

5. **Simplify the expression:**
Now, we can simplify the expression:
\[ \text{LHS} = \frac{\sin i \cdot (1 - \cos i)}{\sin^2 i} \]

6. **Cancel \(\sin i\):**
Since \(\sin^2 i = \sin i \cdot \sin i\), we can cancel one \(\sin i\) from the numerator and denominator:
\[ \text{LHS} = \frac{1 - \cos i}{\sin i} \]

7. **Conclusion:**
Now we have shown that:
\[ \text{LHS} = \frac{1 - \cos i}{\sin i} = \text{RHS} \]
Therefore, we can conclude that:
\[ \frac{\sin i}{1 + \cos i} = \frac{1 - \cos i}{\sin i} \]
Hence, proved.