If `A = 30^(@)`, show that :
(i) `(1 + sin 2A + cos 2A)/(sin A + cos A) = 2 cos A`
(ii) `(cos^(3)A - cos 3A)/(cos A) + (sin^(3)A + sin 3A)/(sin A) = 3`

Answer

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (i): Show that
\[ \frac{1 + \sin 2A + \cos 2A}{\sin A + \cos A} = 2 \cos A \]
where \( A = 30^\circ \).

**Step 1: Substitute \( A \) with \( 30^\circ \)**

We know:
- \( \sin 2A = \sin 60^\circ = \frac{\sqrt{3}}{2} \)
- \( \cos 2A = \cos 60^\circ = \frac{1}{2} \)
- \( \sin A = \sin 30^\circ = \frac{1}{2} \)
- \( \cos A = \cos 30^\circ = \frac{\sqrt{3}}{2} \)

Substituting these values into the expression:

\[ \frac{1 + \frac{\sqrt{3}}{2} + \frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} \]

**Step 2: Simplify the numerator and denominator**

Numerator:
\[ 1 + \frac{\sqrt{3}}{2} + \frac{1}{2} = 1 + \frac{1 + \sqrt{3}}{2} = \frac{2 + 1 + \sqrt{3}}{2} = \frac{3 + \sqrt{3}}{2} \]

Denominator:
\[ \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \]

**Step 3: Substitute back into the expression**

Now substituting back into the expression gives:

\[ \frac{\frac{3 + \sqrt{3}}{2}}{\frac{1 + \sqrt{3}}{2}} = \frac{3 + \sqrt{3}}{1 + \sqrt{3}} \]

**Step 4: Rationalize the denominator**

To simplify \(\frac{3 + \sqrt{3}}{1 + \sqrt{3}}\), we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(3 + \sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{(3 + \sqrt{3} - 3\sqrt{3} - 3)}{1 - 3} = \frac{-2\sqrt{3}}{-2} = \sqrt{3} \]

**Step 5: Calculate \(2 \cos A\)**

Now we calculate \(2 \cos A\):

\[ 2 \cos 30^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \]

**Conclusion for Part (i):**

Thus, we have shown that:

\[ \frac{1 + \sin 2A + \cos 2A}{\sin A + \cos A} = 2 \cos A \]

### Part (ii): Show that
\[ \frac{\cos^3 A - \cos 3A}{\cos A} + \frac{\sin^3 A + \sin 3A}{\sin A} = 3 \]

**Step 1: Substitute \( A \) with \( 30^\circ \)**

We know:
- \( \cos 3A = \cos 90^\circ = 0 \)
- \( \sin 3A = \sin 90^\circ = 1 \)

Substituting these values into the expression:

\[ \frac{\cos^3 30^\circ - 0}{\cos 30^\circ} + \frac{\sin^3 30^\circ + 1}{\sin 30^\circ} \]

**Step 2: Calculate \( \cos^3 30^\circ \) and \( \sin^3 30^\circ \)**

Calculating the values:
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)

Now substituting these values:

\[ \frac{\left(\frac{\sqrt{3}}{2}\right)^3}{\frac{\sqrt{3}}{2}} + \frac{\left(\frac{1}{2}\right)^3 + 1}{\frac{1}{2}} \]

**Step 3: Simplify each term**

First term:
\[ \frac{\frac{3\sqrt{3}}{8}}{\frac{\sqrt{3}}{2}} = \frac{3\sqrt{3}}{8} \cdot \frac{2}{\sqrt{3}} = \frac{3 \cdot 2}{8} = \frac{6}{8} = \frac{3}{4} \]

Second term:
\[ \frac{\frac{1}{8} + 1}{\frac{1}{2}} = \frac{\frac{1 + 8}{8}}{\frac{1}{2}} = \frac{\frac{9}{8}}{\frac{1}{2}} = \frac{9}{8} \cdot 2 = \frac{18}{8} = \frac{9}{4} \]

**Step 4: Combine the results**

Now adding both terms:

\[ \frac{3}{4} + \frac{9}{4} = \frac{12}{4} = 3 \]

**Conclusion for Part (ii):**

Thus, we have shown that:

\[ \frac{\cos^3 A - \cos 3A}{\cos A} + \frac{\sin^3 A + \sin 3A}{\sin A} = 3 \]