Prove that :
(i) `((tan 60^(@)+1)/(tan 60^(@) - 1))^(2) = (1 + cos 30^(@))/(1 - cos 30^(@))`
(ii) `3 "cosec"^(2) 60^(@) - 2 cot^(2) 30^(@) + sec^(2) 45^(@) = 0`

Answer

Let's solve the given problems step by step.

### Part (i): Prove that
\[ \left(\frac{\tan 60^\circ + 1}{\tan 60^\circ - 1}\right)^{2} = \frac{1 + \cos 30^\circ}{1 - \cos 30^\circ} \]

**Step 1: Calculate \(\tan 60^\circ\)**
We know that:
\[ \tan 60^\circ = \sqrt{3} \]

**Step 2: Substitute \(\tan 60^\circ\) into the left-hand side (LHS)**
Substituting the value of \(\tan 60^\circ\):
\[ \text{LHS} = \left(\frac{\sqrt{3} + 1}{\sqrt{3} - 1}\right)^{2} \]

**Step 3: Simplify the expression**
We can simplify the fraction:
\[ \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]
Thus,
\[ \text{LHS} = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \]

**Step 4: Calculate \(\cos 30^\circ\)**
We know that:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

**Step 5: Substitute \(\cos 30^\circ\) into the right-hand side (RHS)**
Substituting the value of \(\cos 30^\circ\):
\[ \text{RHS} = \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}} = \frac{\frac{2 + \sqrt{3}}{2}}{\frac{2 - \sqrt{3}}{2}} = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \]

**Step 6: Rationalize the RHS**
To simplify:
\[ \text{RHS} = \frac{(2 + \sqrt{3})^2}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{4 + 4\sqrt{3} + 3}{4 - 3} = 7 + 4\sqrt{3} \]

**Step 7: Conclusion**
Since LHS = RHS, we have:
\[ \left(\frac{\tan 60^\circ + 1}{\tan 60^\circ - 1}\right)^{2} = \frac{1 + \cos 30^\circ}{1 - \cos 30^\circ} \]
Thus, the first part is proved.

### Part (ii): Prove that
\[ 3 \csc^2 60^\circ - 2 \cot^2 30^\circ + \sec^2 45^\circ = 0 \]

**Step 1: Calculate \(\csc 60^\circ\)**
We know that:
\[ \csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}} \quad \text{(since } \sin 60^\circ = \frac{\sqrt{3}}{2}\text{)} \]
Thus,
\[ \csc^2 60^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]

**Step 2: Calculate \(\cot 30^\circ\)**
We know that:
\[ \cot 30^\circ = \frac{1}{\tan 30^\circ} = \sqrt{3} \]
Thus,
\[ \cot^2 30^\circ = (\sqrt{3})^2 = 3 \]

**Step 3: Calculate \(\sec 45^\circ\)**
We know that:
\[ \sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2} \]
Thus,
\[ \sec^2 45^\circ = (\sqrt{2})^2 = 2 \]

**Step 4: Substitute values into the equation**
Now substituting these values into the equation:
\[ 3 \left(\frac{4}{3}\right) - 2(3) + 2 = 0 \]

**Step 5: Simplify the equation**
Calculating:
\[ 4 - 6 + 2 = 0 \]
Thus, we have:
\[ 0 = 0 \]

**Step 6: Conclusion**
Since LHS = RHS, we have:
\[ 3 \csc^2 60^\circ - 2 \cot^2 30^\circ + \sec^2 45^\circ = 0 \]
Thus, the second part is proved.