Question

tan² theta = 1- e² the prove that sec theta + tan³theta × cosec theta = (2- e²)²/³​

Answer 1

$\huge\underline\bold\red{★Answer★}$

$given \: \: \: {tan}^{2} \theta = 1 - {e}^{2} \\ lhs \\ = sec \theta + {tan}^{3} \theta cosec \theta \\ = \sqrt{1 + {tan}^{2} \theta } + {tan}^{2} \theta \: tan \theta \: ( \sqrt{1 + {cot}^{2} \theta } \: \: ) \\ = \sqrt{1 + {tan}^{2} \theta } + {tan}^{2} \theta \: tan \theta \ \sqrt{1 + { \frac{1}{tan ^{2} \theta } }} \\ = \sqrt{1 + 1 - {e}^{2} } + (1 - {e}^{2} )( \sqrt{1 - {e}^{2} } \: )\sqrt{1 + \frac{1}{1 - {e}^{2} } } \: \\ = \sqrt{2 - {e}^{2} } + (1 - {e}^{2} ) \sqrt{1 - {e}^{2} } \sqrt{ \frac{1 - {e}^{2} + 1 }{1 - {e}^{2} } } \\ = \sqrt{2 - {e}^{2} } +( 1 - {e}^{2} ) \sqrt{1 - {e}^{2} } ( \frac{ \sqrt{2 - {e}^{2} } }{ \sqrt{1 - {e}^{2} } } ) \\ = \sqrt{2 - {e}^{2} } + (1 - {e}^{2} ) \sqrt{2 - {e}^{2} } \\ = \sqrt{2 - {e}^{2} } (1 + 1 - {e}^{2} ) \\ = \sqrt{2 - {e}^{2} } (2 - {e}^{2} ) \\ = (2 - {e}^{2} ) ^{ \frac{1}{2} + 1} \\ = (2 - {e}^{2} ) ^{ \frac{3}{2} }$