Question

plz slove this question ​

Answer 1

Step-by-step explanation:

$\frac{ {tan}^{3} \alpha - 1}{tan \alpha - 1} \\ \\ = \frac{(tan \alpha - 1)( {tan}^{2} \alpha + tan \alpha + 1) }{(tan \alpha - 1)} \\ \\ = {tan}^{2} \alpha + tan \alpha + 1$

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