Question

Can anyone please answer this question.
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Answer 1

Answer:

$({\frac{1 - {tan}^{2}α }{1 + {tan}^{2}α })}^{2} + { (\frac{2 \:tan \: α}{1 + {tan}^{2}α } )}^{2} = 1$

Step-by-step explanation:

$3 \: sin \: α = 2 \: cos \: α \\ \frac{sin \: α}{cos \: α} = \frac{2}{3} \\ tan \: α = \frac{2}{3} \\ ({\frac{1 - {tan}^{2}α }{1 + {tan}^{2}α })}^{2} + { (\frac{2 \:tan \: α}{1 + {tan}^{2}α } )}^{2} =1\\ { (\frac{1 - \frac{4}{9} }{1 + \frac{4}{9} } )}^{2} + { (\frac{2 \times \frac{2}{3} }{1 + \frac{4}{9} }) }^{2} =1\\ {( \frac{ \frac{9 - 4}{9} }{ \frac{9 + 4}{9} } )}^{2} + { (\frac{2 \times \frac{2}{3} }{ \frac{9 + 4}{9} } )}^{2} =1\\ {( \frac{ \frac{5}{9} }{ \frac{13}{9} }) }^{2} + { (\frac{ \frac{4}{3} }{ \frac{13}{9} } )}^{2} =1 \\ { (\frac{5}{9} \times \frac{9}{13} )}^{2} + { (\frac{4}{3} \times \frac{9}{13} )}^{2} =1 \\ { (\frac{5}{13}) }^{2} + { (\frac{12}{13} )}^{2} =1\\ \frac{25}{169} + \frac{144}{169} =1 \\ \frac{25 + 144}{169} =1\\ \frac{169}{169} \\ = 1$\\LHS=RHS\\Hence proved