Show that
(i) `sin^(8)A-cos^(8)A=(sin^(2)A-cos^(2)A)(1-2sin^(2)A.cos^(2)A)`
(ii) `(1)/(sec A-tan A)-(1)/(cos A)=(1)/(cos A)-(1)/(sec A + tan A)`

Answer

To solve the given problems, we will break down each part step by step.

### Part (i): Show that
\[ \sin^8 A - \cos^8 A = (\sin^2 A - \cos^2 A)(1 - 2\sin^2 A \cos^2 A) \]

#### Step 1: Start with the left-hand side (LHS)
We begin with the expression:
\[ \sin^8 A - \cos^8 A \]

#### Step 2: Use the difference of squares
Recognizing that \(a^2 - b^2 = (a - b)(a + b)\), we can rewrite:
\[ \sin^8 A - \cos^8 A = (\sin^4 A - \cos^4 A)(\sin^4 A + \cos^4 A) \]

#### Step 3: Apply the difference of squares again
Now, we apply the difference of squares to \(\sin^4 A - \cos^4 A\):
\[ \sin^4 A - \cos^4 A = (\sin^2 A - \cos^2 A)(\sin^2 A + \cos^2 A) \]
Since \(\sin^2 A + \cos^2 A = 1\), we have:
\[ \sin^4 A - \cos^4 A = (\sin^2 A - \cos^2 A)(1) \]
Thus,
\[ \sin^8 A - \cos^8 A = (\sin^2 A - \cos^2 A)(\sin^4 A + \cos^4 A) \]

#### Step 4: Simplify \(\sin^4 A + \cos^4 A\)
We know:
\[ \sin^4 A + \cos^4 A = (\sin^2 A + \cos^2 A)^2 - 2\sin^2 A \cos^2 A = 1 - 2\sin^2 A \cos^2 A \]
So, substituting this back, we get:
\[ \sin^8 A - \cos^8 A = (\sin^2 A - \cos^2 A)(1 - 2\sin^2 A \cos^2 A) \]

#### Conclusion for Part (i)
Thus, we have shown that:
\[ \sin^8 A - \cos^8 A = (\sin^2 A - \cos^2 A)(1 - 2\sin^2 A \cos^2 A) \]

---

### Part (ii): Show that
\[ \frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A} \]

#### Step 1: Start with the left-hand side (LHS)
We begin with:
\[ \frac{1}{\sec A - \tan A} - \frac{1}{\cos A} \]

#### Step 2: Rewrite \(\sec A\) and \(\tan A\)
Recall that:
\[ \sec A = \frac{1}{\cos A}, \quad \tan A = \frac{\sin A}{\cos A} \]
Thus:
\[ \sec A - \tan A = \frac{1 - \sin A}{\cos A} \]
So, we rewrite the LHS:
\[ \frac{1}{\frac{1 - \sin A}{\cos A}} - \frac{1}{\cos A} = \frac{\cos A}{1 - \sin A} - \frac{1}{\cos A} \]

#### Step 3: Take the common denominator
The common denominator for the two fractions is \(\cos A(1 - \sin A)\):
\[ \frac{\cos^2 A - (1 - \sin A)}{\cos A(1 - \sin A)} = \frac{\cos^2 A + \sin A - 1}{\cos A(1 - \sin A)} \]
Using the identity \(\cos^2 A = 1 - \sin^2 A\):
\[ \cos^2 A + \sin A - 1 = (1 - \sin^2 A) + \sin A - 1 = -\sin^2 A + \sin A \]
Thus, we have:
\[ \frac{\sin A(1 - \sin A)}{\cos A(1 - \sin A)} = \frac{\sin A}{\cos A} = \tan A \]

#### Step 4: Now simplify the right-hand side (RHS)
The RHS is:
\[ \frac{1}{\cos A} - \frac{1}{\sec A + \tan A} \]
Again, rewrite \(\sec A + \tan A\):
\[ \sec A + \tan A = \frac{1 + \sin A}{\cos A} \]
Thus, we have:
\[ \frac{1}{\cos A} - \frac{\cos A}{1 + \sin A} = \frac{1(1 + \sin A) - \cos^2 A}{\cos A(1 + \sin A)} \]
Using \(\cos^2 A = 1 - \sin^2 A\):
\[ 1(1 + \sin A) - (1 - \sin^2 A) = \sin A + \sin^2 A \]
So, we have:
\[ \frac{\sin A(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{\sin A}{\cos A} = \tan A \]

#### Conclusion for Part (ii)
Thus, we have shown that:
\[ \frac{1}{\sec A - \tan A} - \frac{1}{\cos A} = \frac{1}{\cos A} - \frac{1}{\sec A + \tan A} \]

---