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Answer 1

$To \: prove : \bigg(\dfrac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta} \bigg) {}^{2} = \dfrac{1 - \cos \theta}{1 + \cos \theta }$



Solving left hand side,

$\implies \bigg( \dfrac{1 + \sin \theta- \cos \theta}{1 + \sin \theta+ \cos \theta} \bigg) {}^{2} \\ \\ \\ \implies \bigg( \dfrac{\sin \theta + 1 - \cos \theta}{ \sin \theta + 1+ \cos \theta} \bigg) {}^{2}$



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From the properties of trigonometry, we know
sin²A = 1 - cos²A
Or, sinA = √{ 1 - cos²A }
Or, sinA = √{ 1 - cosA }{ 1 + cosA }
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$\implies \bigg(\dfrac{ \sqrt{ 1 - \cos {}^{2} \theta} + (1- \cos \theta) }{ \sqrt{1 - \cos {}^{2} \theta } +( 1+ \cos \theta)} \bigg) {}^{2} \\ \\ \\ \implies \bigg(\dfrac{ \sqrt{ 1 - \cos {}^{2} \theta } + \sqrt{ (1- \cos \theta) {}^{2} } }{ \sqrt{1 - \cos {}^{2} \theta } + \sqrt{ ( 1+ \cos \theta) {}^{2} }} \bigg) {}^{2} \\ \\ \\ \implies \bigg( \dfrac{ \sqrt{ (1 - \cos {}^{} \theta)(1 + \cos \theta) } + \sqrt{ (1- \cos \theta) {}^{2} } }{ \sqrt{(1 - \cos \theta)(1 + \cos \theta) } + \sqrt{ ( 1+ \cos \theta) {}^{2} }} \bigg) {}^{2} \\ \\ \\ \implies \bigg(\dfrac{\sqrt{1 - \cos \theta} \{ \sqrt{1 + \cos \theta} + \sqrt{1 - \cos \theta \}} }{ \sqrt{1 + \cos \theta }\{ \sqrt{1 - \cos \theta} + \sqrt{1 + \cos \theta }\}} \bigg) {}^{2} \\ \\ \\ \implies \dfrac{1 - \cos \theta}{1 + \cos \theta}$

Hence, proved.

Answer 2

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