Prove that: sec⁶x – tan⁶x = 1 + 3sec²x × tan²x

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: LHS simplifies to RHS, as shown below.

Let \(a=\sec^2x\), \(b=\tan^2x\). Then \(a-b=1\).

Using \(a^3-b^3=(a-b)^3+3ab(a-b)\):

\[\sec^6x-\tan^6x=1^3+3ab(1)\]

\[=1+3\sec^2x\tan^2x=\text{RHS}\]

Doubtly AI · 24 Oct 2026 AI-verified · Oct 2026 Report

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