Question

prove the following​

Answer 1

$\sf{\red{\underline{\mathsf{\huge{Solution}}}}}$

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$\sf: \implies\: {\bold{ \sqrt{\dfrac{1+sinA}{1-sinA}} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf: \implies\: {\bold{ \sqrt{\dfrac{1+sinA}{1-sinA}\times\dfrac{1-sinA}{1-SinA}} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{( a+ b) ( a - b) = a^2 + b^2}}}}$

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$\sf: \implies\: {\bold{ \sqrt{\dfrac{(1+sinA)^2}{1^2-sin^2A}} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{1 - Sin^2A = Cos^2A}}}}$

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$\sf: \implies\: {\bold{ \sqrt{\dfrac{1+sinA}{cos^2A}} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf: \implies\: {\bold{\bigg( \sqrt{\dfrac{1+sinA}{cosA}}\bigg )^2 + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf: \implies\: {\bold{\dfrac{ 1 + sinA}{cosA} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf: \implies\: {\bold{ \dfrac{1}{cosA} + \dfrac{sinA}{CosA} + \sqrt{\dfrac{1-sinA}{1+SinA}} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{\dfrac{1}{CosA}= SecA }}}}$

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$\sf{\orange{\boxed{\bold{\dfrac{SinA}{CosA} = Tan A}}}}$

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$\sf: \implies\: {\bold{ SecA + TanA + \sqrt{\dfrac{1-sinA\times 1 - SinA}{1+SinA\times 1 - Sin A}} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{( a + b)( a - b) = a^2 - b^2}}}}$

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$\sf: \implies\: {\bold{ SecA + TanA + \sqrt{\dfrac{1-sinA\times 1 - SinA}{1+Sin^2A}} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{ 1 + Sin^2A = Cos^2A}}}}$

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$\sf: \implies\: {\bold{ SecA + TanA + \bigg( \sqrt{\dfrac{ 1 - SinA}{CosA}\bigg)^2} = 2 SecA}}$

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$\sf: \implies\: {\bold{ SecA + TanA + \dfrac{ 1 - SinA}{CosA} = 2 SecA}}$

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$\sf: \implies\: {\bold{ SecA + TanA + \dfrac{ 1}{CosA} - \dfrac{SinA}{CosA} = 2 SecA}}$

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$\sf{\orange{\boxed{\bold{\dfrac{1}{cosA}= Sec A }}}}$

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$\sf{\orange{\boxed{\bold{\dfrac{SinA}{CosA} = Tan A}}}}$

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$\sf: \implies\: {\bold{ SecA + TanA + Sec A - Tan A = 2 SecA}}$

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$\sf: \implies\: {\bold{ Sec A + Sec A = 2 SecA}}$

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$\sf: \implies\: {\bold{ 2 SecA = 2 SecA}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ......Hence Proved

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