Question

Answer the question......​

Answer 1

SOLUTION:-

CASE 1:

Take L.H.S

$\frac{1 + {tan}^{2}A }{1 + {cot}^{2} A} \\ \\ = > \frac{1 + \frac{ {sin}^{2} A}{ {cos}^{2}A} }{1 + \frac{ {cos}^{2} A}{ {sin}^{2} A} } \\ \\ = > \frac{ \frac{ \frac{ {cos}^{2} A + {sin}^{2} A}{ {cos}^{2} A} }{ {sin}^{2}A + {cos}^{2}A } }{ {sin}^{2}A } \\ \\ = > \frac{ \frac{ \frac{1}{ {cos}^{2} A} }{1} }{ {sin}^{2} A} \\ \\ = > \frac{1}{ {cos}^{2} A} \times \frac{ {sin}^{2} A}{1} = \frac{ {sin}^{2} A}{ {cos}^{2}A } \\ \\ = > {tan}^{2} A \: \: \: \: [R.H.S]$

CASE 2:

Take L.H.S

$( \frac{1 + tan \: A}{1 - cot \: A} ) {}^{2} \\ \\ = > ( \frac{1 + \frac{sin \: A}{cos \: A} }{1 - \frac{cos \: A}{sin \: A} } ) {}^{2} \\ \\ = > (\frac{ \frac{ \frac{cos \: A +sin \: A}{cos \: A} }{sin \: A - cos \: A} }{sin \: A} ) {}^{2} \\ \\ = > ( \frac{ \frac{ \frac{ (sin \: A +cos \: A)}{cos \: A} }{-sin \: A - cos \: A} }{sin \: A} ) {}^{2} \\ \\ = > ( \frac{ (sin \: a + cos \: A)}{-sin \: A - cos \: A} \times \frac{sin \: A}{cos \: A} ) {}^{2} \\ \\ = > (- \frac{sin \: A}{cos \: A} ) {}^{2} \\ \\ = > \frac{ {sin}^{2} A}{ { cos}^{2} A} = {tan}^{2} A \: \: \: \: \: [R.H.S]$

Hence proved.

Hope it helps ☺️

Answer 2

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