Question

show that √1+sin∆÷1-sin∆=sec∆+tan∆​

Answer 1

TO PROVE :–

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

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SOLUTION :–

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• Let's take L.H.S. –

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

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• Now 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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• We should write this as –

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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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$\bf \: \: = R.H.S.$

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$\bf \: \: \: \: \: \: \: { \underbrace{ Hence \: \: proved}}$