Question

prove that
please answer
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Answer 1

Answer:

Hey mate....

please refer the above attachment!

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Answer 2

$\huge{\bold{ To Prove :-}}$

$\frac{1 - {tan}^{2}A }{1 + {tan}^{2} A} = 1 - 2 {sin}^{2} A\\ \\ {\purple{\boxed{\large{\bold{Trigonomertry \: formula \: used}}}}} \\ \\ 1. \: {tan}^{2} \theta = \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta } \\ \\ 2. \: {sin}^{2} \theta+ {cos}^{2} \theta = 1 \\ \\ 3. \: {cos}^{2} \theta = 1 - {sin}^{2} \theta \\ \\ \huge{\bold{ Proof:-}} \\ \\ \small{\bold{ LHS}} \\ \\ = \frac{1 - {tan}^{2}A }{1 + {tan}^{2}A } \\ \\ = \frac{1 - \frac{ {sin}^{2} A}{ {cos}^{2}A } }{1 + \frac{ {sin}^{2}A }{ {cos}^{2}A } } \\ \\ = \frac{ \frac{ {cos}^{2}A - {sin}^{2}A }{ {cos}^{2} A} }{ \frac{ {cos}^{2}A + {sin}^{2}A }{ {cos}^{2}A } } \\ \\ = \frac{ {cos}^{2}A - {sin}^{2}A }{ {cos}^{2}A + {sin}^{2} A} \\ \\ = {cos}^{2} A - {sin}^{2} A \\ \\ = 1 - {sin}^{2} A- {sin}^{2} A \\ \\ = 1 - 2 {sin}^{2} A = RHS \\ \\ therefore \: \\ {\purple{\boxed{\large{\bold{LHS = RHS}}}}} \\ \\ \huge{\bold{ Hence \: proved}}$

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