Question

prove that:(as attached)
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Answer 1

Step-by-step explanation:

We have,

$\frac{1 + \tan^{3} ( \theta) }{1 + \tan( \theta) } + \tan( \theta) - \sec^{2} ( \theta)$

$= \frac{(1 + \tan( \theta) )(1 + \tan^{2} ( \theta) - \tan( \theta) }{1 + \tan( \theta) } + \tan( \theta) - \sec^{2} ( \theta)$

$= 1 + \tan^{2} ( \theta) - \tan( \theta) + \tan( \theta) - \sec^{2} ( \theta)$

$= \sec^{2} ( \theta)- \sec^{2} ( \theta)$

$= 0$