Question

Prove that √1 + sin triangle ÷ 1 - sin triangle = sec Triangle + tan Triangle ​

Answer 1

${ \bf{ \underline{ \blue{ \underline{ \blue{ Given}}}} : - }}$

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$\bf \implies \sqrt{ \dfrac{1 + \sin( \triangle) }{1 - \sin( \triangle) }} = \sec( \triangle) + \tan( \triangle)$

${ \bf{ \underline{ \blue{ \underline{ \blue{ To- Prove }}}} : - }}$

$\bf \implies \sqrt{ \dfrac{1 + \sin( \triangle) }{1 - \sin( \triangle) }} = \sec( \triangle) + \tan( \triangle)$

${ \bf{ \underline{ \blue{ \underline{ \blue{ Solution }}}} : - }}$

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L.H.S

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$\bf \: \: = \sqrt{ \dfrac{1 + \sin( \triangle) }{1 - \sin( \triangle) }}$

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( Rationalization of denominator )

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$\bf \: \: = \sqrt{ \dfrac{1 + \sin( \triangle) }{1 - \sin( \triangle) } \times \dfrac{1 + \sin( \triangle) }{1 + \sin( \triangle)}}$

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$\bf \: \: = \sqrt{ \dfrac{ \{1 + \sin( \triangle) \}^{2} }{ \{1 - \sin( \triangle) \} \{1 + \sin( \triangle)\} }}$

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$\bf \: \: = \sqrt{ \dfrac{\{1 + \sin( \triangle) \}^{2} }{ {(1)}^{2} - \sin^{2} ( \triangle)}}$

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$\bf \: \: = \sqrt{ \dfrac{\{1 + \sin( \triangle) \}^{2}}{1 - \sin^{2} ( \triangle)}}$

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$\bf \: \: = \sqrt{ \dfrac{\{1 + \sin( \triangle) \}^{2}}{\cos^{2} ( \triangle)}}$

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$\bf \: \: = \dfrac{1 + \sin( \triangle) }{\cos ( \triangle)}$

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$\bf \: \: = \dfrac{1}{\cos ( \triangle)} + \dfrac{\sin( \triangle) }{\cos ( \triangle)}$

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$\bf \: \: = \sec ( \triangle)+ \tan( \triangle)$

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= R.H.S

L.H.S = R.H.S

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$\small{\underline{\sf{\blue{Hence-}}}}$

( Proved)

Answer 2

answer : R.H.S

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