(i) `cos20^(@) + cos100^(@) + cos 140^(@) =0`
(ii) `sin 50^(@) - sin 70^(@) + sin 10^(@) =0`
(iii) `2 cos "" ( pi )/( 13) cos "" ( 9 pi )/( 13) + cos "" (3pi)/( 13) + cos "" ( 5pi )/( 13) =0`

Answer

Let's solve the three parts step by step.

### Part (i): Prove that \( \cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0 \)

1. **Use the Cosine Addition Formula**:
We can use the formula for the sum of cosines:
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \]
Set \( A = 20^\circ \) and \( B = 100^\circ \):
\[ \cos 20^\circ + \cos 100^\circ = 2 \cos\left(\frac{20^\circ + 100^\circ}{2}\right) \cos\left(\frac{20^\circ - 100^\circ}{2}\right) \]

2. **Calculate the Averages**:
\[ \frac{20^\circ + 100^\circ}{2} = 60^\circ, \quad \frac{20^\circ - 100^\circ}{2} = -40^\circ \]
Thus,
\[ \cos 20^\circ + \cos 100^\circ = 2 \cos(60^\circ) \cos(-40^\circ) \]

3. **Substitute Values**:
We know \( \cos(60^\circ) = \frac{1}{2} \):
\[ \cos 20^\circ + \cos 100^\circ = 2 \cdot \frac{1}{2} \cdot \cos(40^\circ) = \cos(40^\circ) \]

4. **Combine with \( \cos 140^\circ \)**:
Now we add \( \cos 140^\circ \):
\[ \cos 20^\circ + \cos 100^\circ + \cos 140^\circ = \cos(40^\circ) + \cos(140^\circ) \]

5. **Use the Cosine Addition Formula Again**:
Now apply the cosine formula again:
\[ \cos(40^\circ) + \cos(140^\circ) = 2 \cos\left(\frac{40^\circ + 140^\circ}{2}\right) \cos\left(\frac{40^\circ - 140^\circ}{2}\right) \]

6. **Calculate the New Averages**:
\[ \frac{40^\circ + 140^\circ}{2} = 90^\circ, \quad \frac{40^\circ - 140^\circ}{2} = -50^\circ \]
Thus,
\[ \cos(40^\circ) + \cos(140^\circ) = 2 \cos(90^\circ) \cos(-50^\circ) \]

7. **Final Value**:
Since \( \cos(90^\circ) = 0 \):
\[ 2 \cdot 0 \cdot \cos(-50^\circ) = 0 \]
Therefore, \( \cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0 \).