Trig1.12

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Prove that \frac{\tan^3 \theta-1}{\tan\theta-1}=\sec^2 \theta+\tan\theta\,

Applying the algebra formula on the LHS,we get

LHS=\frac{\tan\theta-1)(\tan^2 \theta+\tan\theta+1)}{\tan\theta-1}\,

\tan^2 \theta+1+\tan\theta\,

\sec^2 \theta+\tan\theta\,=RHS


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