Prove that:
(i) `sec^(4)A-sec^(2)A="tan"^(4)A+"tan"^(2)A`
(ii) `"sin"^(8)theta-cos^(8)theta=("sin"^(2)theta-cos^(2)theta)(1-2"sin"^(2)thetacos^(2)theta)`.

Answer

Let's solve the given problems step by step.

### Part (i): Prove that \( \sec^4 A - \sec^2 A = \tan^4 A + \tan^2 A \)

**Step 1:** Start with the left-hand side (LHS):
\[ \sec^4 A - \sec^2 A \]

**Step 2:** Factor out \( \sec^2 A \) from the LHS:
\[ \sec^2 A (\sec^2 A - 1) \]

**Step 3:** Use the identity \( \sec^2 A - 1 = \tan^2 A \):
\[ \sec^2 A \tan^2 A \]

**Step 4:** Rewrite \( \sec^2 A \) in terms of \( \tan^2 A \):
\[ \sec^2 A = 1 + \tan^2 A \]

**Step 5:** Substitute this back into the equation:
\[ (1 + \tan^2 A) \tan^2 A \]

**Step 6:** Distribute \( \tan^2 A \):
\[ \tan^2 A + \tan^4 A \]

**Step 7:** Rearrange the terms:
\[ \tan^4 A + \tan^2 A \]

**Conclusion:** The left-hand side equals the right-hand side:
\[ \sec^4 A - \sec^2 A = \tan^4 A + \tan^2 A \]
Thus, the equation is proved.