Prove that: (tan³θ – 1)/(tanθ – 1) = sec²θ + tanθ

Doubtly Asked 24 Oct 2026 Save WhatsApp Report

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Answer: LHS simplifies to RHS, as shown below.

Using \(a^3-1=(a-1)(a^2+a+1)\) with \(a=\tan\theta\):

\[\text{LHS}=\frac{(\tan\theta-1)(\tan^2\theta+\tan\theta+1)}{\tan\theta-1}\]

\[=(1+\tan^2\theta)+\tan\theta=\sec^2\theta+\tan\theta=\text{RHS}\]

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

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