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प्रश्न
Prove the following.
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उत्तर
\[\frac{\tan^3 \theta - 1}{\tan\theta - 1}\]
\[ = \frac{\left( \tan\theta - 1 \right)\left( \tan^2 \theta + \tan\theta \times 1 + 1 \right)}{\tan\theta - 1} \left[ a^3 - b^3 = \left( a - b \right)\left( a^2 + ab + b^2 \right) \right]\]
\[ = \tan^2 \theta + \tan\theta + 1\]
\[ = \sec^2 \theta + \tan\theta \left( 1 + \tan^2 \theta = \sec^2 \theta \right)\]
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