Question

please solved this:-----if tanθ=a/b then sinθ/cos^3θ + cosθ/sin^3θ=??​

Answer 1

$tan \theta = \frac{a}{b} \\ sec \theta = \sqrt{1 + {tan}^{2} \theta } \\ = \sqrt{1 + \frac{ {a}^{2} }{ {b}^{2} } } \\ = \frac{ \sqrt{ {a}^{2} + {b}^{2} } }{b} \\ \\cos \theta = \frac{b}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ sin \theta = \sqrt{1 - {cos}^{2} \theta } \\ = \sqrt{1 - \frac{ {b}^{2} }{ {a}^{2} + {b}^{2} } } \\ = \frac{a}{ \sqrt{ {a}^{2} + {b}^{2} } } \\ \\ \huge Now \: \: \: \: \: \frac{sin \theta}{ {cos}^{3} \theta} + \frac{cos \theta}{ {sin}^{3} \theta} \\ \\ = \frac{ {sin}^{4} \theta + {cos}^{4} \theta}{ {sin}^{3} \theta {cos}^{3} \theta}\\ \\ = \frac{( { {sin}^{2} \theta + {cos}^{2} \theta })^{2} - 2 {sin}^{2} \theta {cos}^{2} \theta }{ {sin}^{3} \theta {cos}^{3} \theta } \\ \\ = \frac{1 - 2 {sin}^{2} \theta \: {cos}^{2} \theta}{ {sin}^{3} \theta \: {cos}^{3} \theta} \\ \\ = \frac{1 - 2. \frac{ {a}^{2} }{ {a}^{2} + {b}^{2} }. \frac{ {b}^{2} }{ {a}^{2} + {b}^{2} } }{ { \frac{ {a}^{3} }{( {a}^{2} + {b}^{2} )\sqrt{ {a}^{2} + {b}^{2} } } }. \frac{ {b}^{3} }{( {a}^{2} + {b}^{2} ) \sqrt{ {a}^{2} + {b}^{2} } } } \\ \\ = \frac{ \frac{( { {a}^{2} + {b}^{2} })^{2} - 2 {a}^{2} {b}^{2} }{( { {a}^{2} + {b}^{2} })^{2} } }{ \frac{ {a}^{3} {b}^{3} }{ ({ {a}^{2} + {b}^{2} })^{3} } } \\ \\ = \frac{ {a}^{4} + {b}^{4} }{( { {a}^{2} + {b}^{2} })^{2} } \times \frac{( { {a}^{2} + {b}^{2} })^{3} }{ {a}^{3} {b}^{3} } \\ \\ = \frac{( {a}^{4} + {b}^{4} )( {a}^{2} + {b}^{2} )}{ {a}^{3} {b}^{3} }$