Question

Prove the following ​

Answer 1

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$\large {\frac{tan³ \theta - 1 }{tan \theta -1 }= sec²\theta + tan \theta}$

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$\large {\frac{tan³ \theta - 1 }{tan \theta -1 }}$

$\mathfrak{\large {\frac{(tan \theta - 1)(tan² \theta + tan \theta + 1²}{tan \theta -1 }}}$

$\large {\frac{\cancel{(tan \theta - 1)}(tan² \theta + tan \theta + 1²}{\cancel{tan \theta -1 }}}$

$\mathfrak{\large {tan² \theta + tan \theta + 1²}}$

$\mathfrak{\large {tan² \theta +1+ tan \theta}}$

$\mathfrak{\large {sec² \theta + tan \theta}}$