Question

Prove that \( \sec^4 \theta - \tan^4 \theta = 1 + 2\tan^2 \theta \).

Hint:

Use algebraic identities like \( a^2 - b^2 = (a+b)(a-b) \) to simplify expressions.

Answer 1

We use the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta. \] Expanding the given expression:
\[ \sec^4 \theta - \tan^4 \theta = (\sec^2 \theta + \tan^2 \theta)(\sec^2 \theta - \tan^2 \theta). \] Since, \[ \sec^2 \theta - \tan^2 \theta = 1. \] \[ \sec^4 \theta - \tan^4 \theta = (\sec^2 \theta + \tan^2 \theta) \times 1. \] \[ = \sec^2 \theta + \tan^2 \theta. \] Using \( \sec^2 \theta = 1 + \tan^2 \theta \), we get:
\[ 1 + 2\tan^2 \theta. \]