Question

please brooos or sisters prove that​

Answer 1

❏QuesTion :-

Prove that :

$\to \: \frac{ \cos A }{ \csc A \: \: + 1} + \frac{ \cos A\: }{ \csc A \: - 1} = 2 \tan A$

➫Used Formula :-

$\boxed{ \star} \: \: \: { \csc}^{2} \theta - { \cot}^{2} \theta = 1 \\ \\ \boxed{ \star} \: \: \: \frac{ \sin \theta}{ \cos \theta} = \tan \theta \\ \\ \boxed{ \star} \: \: \frac{ \cos \theta}{ \sin \theta} = \cot \theta \:$

➤Solution :-

Taking Left hand side ( LHS)

$\to \frac{ \cos A }{ \csc A \: \: + 1} + \frac{ \cos A\: }{ \csc A \: - 1} \\ \\ \sf \: taling \: lcm \\ \\ \to \frac{ \cos A( \csc A - 1) + \cos A ( \csc A + 1)}{( \csc A \: \: + 1)( \csc A - 1)} \\ \: \\ \: \: \: \: \: \: \: \: \boxed{ \because \: (x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \\ \to \frac{\cos A \csc A \: - \cos A +\cos A \csc A +\cos A }{ { \csc}^{2} A - 1} \\ \because \: { \csc}^{2} \theta - 1 = { \cot}^{2} \theta \\ \therefore \\ \to \frac{2 \: \csc A \: \cos A}{ { \cot}^{2} A} \\ \because \csc \theta = \frac{1}{ \sin \theta} \: and \: \cot \theta = \frac{ \cos \theta}{ \sin \theta} \\ \therefore \\ \\ \to \: 2 \frac{ \cos A}{ \sin A \: \times \frac{ { \cos }^{2} A}{ {\sin }^{2} A } } \\ \\ \to \: 2 \frac{1}{ \frac{\cos A}{ \sin A } } \\ \\ \to \: 2 \frac{\sin A}{ \cos A\: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\because \: \frac{ \sin \theta}{ \cos \theta} = \tan \theta} \\ \\ \to \: 2 \tan A \\ \\ \to \boxed{ \frac{ \cos A }{ \csc A \: \: + 1} + \frac{ \cos A\: }{ \csc A - 1} = 2 \tan A }\\ \:$

Hence proved .

Answer 2

$\huge\underline\mathrm{Question-}$

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Prove that : $\tt{\dfrac{CosA}{CosecA+1}+\dfrac{CosA}{CosecA-1}}$ = $\tt{2\:TanA}$

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$\huge\underline\mathrm{Solution-}$

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Taking LHS,

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$\tt{\dfrac{CosA}{CosecA+1}+\dfrac{CosA}{CosecA-1}}$

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By taking LCM,

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$\mapsto$ $\tt{\dfrac{(CosecA-1)(CosA)+(CosA)(CosecA+1)}{(CosecA+1)(CosecA-1)}}$

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By using : (a+b)(a-b) = a² - b²

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$\mapsto$ $\tt{\dfrac{CosecACosA-\cancel{CosA}+CosACosecA+\cancel{CosA}}{Cosec^2\:A-1}}$

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By using : Cosec²A - 1 = Cot²A

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$\mapsto$ $\tt{\dfrac{2CosecACotA}{Cot^2\:}}$

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By changing in the Ratios of sin and cos.

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$\mapsto$ $\tt{\dfrac{\frac{ 2\cos(A) }{ \sin(A) }}{ { \cot(A) }^{2}}}$

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We know that, $\tt{\bold{\dfrac{CosA}{SinA}=CotA}}$ and $\tt{\bold{\dfrac{1}{CotA}=TanA}}$

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$\mapsto$ $\tt{\dfrac{2\cancel{CotA}}{\cancel{(CotA)}(CotA)}}$

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$\mapsto$ $\tt{2TanA}$

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$\mapsto$ RHS

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$\large{\boxed{\tt{\therefore\:\dfrac{CosA}{CosecA+1}+\dfrac{CosA}{CosecA-1}=2TanA}}}$

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Hence proved!