Prove that \(\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta\).
Identity holds true.
Step 1: Recall the Pythagorean trigonometric identity:
\(\sec^2\theta = 1 + \tan^2\theta\).
Step 2: Start with the RHS (Right Hand Side):
\(\sec^4\theta - \sec^2\theta\).
Step 3: Factorize the RHS:
\(\sec^4\theta - \sec^2\theta = \sec^2\theta(\sec^2\theta - 1)\).
Step 4: Use the identity \(\sec^2\theta - 1 = \tan^2\theta\):
So, RHS = \(\sec^2\theta \times \tan^2\theta\).
Step 5: Replace \(\sec^2\theta\) using the identity:
\(\sec^2\theta = 1 + \tan^2\theta\).
Step 6: Substitute into RHS:
RHS = \((1 + \tan^2\theta) \times \tan^2\theta\).
Step 7: Expand the product:
RHS = \(\tan^2\theta + \tan^4\theta\).
Step 8: Observe that this is exactly the LHS (Left Hand Side):
\(\tan^4\theta + \tan^2\theta\).
Conclusion: Since RHS = LHS, the given identity is proved.