Prove that sec⁴ A - sec² A = tan⁴ A + tan² A, and find both sides at A = 60°.

Step 1. Factorise. $\sec^4 A - \sec^2 A = \sec^2 A(\sec^2 A - 1)$.

Step 2. Use the identity twice. $\sec^2 A = 1 + \tan^2 A$ and $\sec^2 A - 1 = \tan^2 A$, so the left side is $(1 + \tan^2 A)\tan^2 A$.

Step 3. Expand. That is $\tan^2 A + \tan^4 A$, the right side.

Step 4. At 60°. $\sec 60^\circ = 2$: $16 - 4 = 12$. $\tan^2 60^\circ = 3$: $9 + 3 = 12$.