Question

√1-cos A / √1+cos A = cosec A - cotA​

Answer 1

√1-cosA/√1+cosA

√1-cosA/√1+cosA×√1-cosA/√1-cosA

√(1-cosA)²/√1-cos²A

1-cosA/√sin²A

1-cosA/sinA

1/sinA-cosA/sinA

cosecA - cotA

LHS= RHS

Answer 2

$\sf{\underline{\underline{\huge{\bold{Solution}}}}}$

${\bold{\sqrt{\dfrac{1 - cos A}{1 + cos A}} = cosec A - cot A}}$

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$\sf{\underline{\underline{\large{\bold{LHS}}}}}$

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$\sf : \implies\:{\bold{\sqrt{\dfrac{1 - cos A}{1 + cos A}}}}$

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• Rationalizing the denominator

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$\sf : \implies\:{\bold{\sqrt{\dfrac{1 - cos A\times 1 - cos A}{1 + cos A\times 1 - cos A}}}}$

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$\sf : \implies\:{\bold{\sqrt{\dfrac{(1 - cos A)^2}{1 + cos A\times 1 - cos A}}}}$

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

⠀⠀⠀⠀

$\sf : \implies\:{\bold{\sqrt{\dfrac{(1 - cos A)^2}{1^2 - cos ^2A}}}}$

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$\sf : \implies\:{\bold{\sqrt{\dfrac{(1 - cos A)^2}{1 - cos ^2A}}}}$

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$\sf{\red{\boxed{\bold{1 - cos^2 a = sin^2 a}}}}$

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$\sf : \implies\:{\bold{\sqrt{\dfrac{(1 - cos A)^2}{sin^2 a}}}}$

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$\sf : \implies\: \bigg( {\bold {\sqrt{\dfrac{1 - cos A}{sin A}}}}\bigg ) ^2$

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$\sf : \implies\:{\bold{{\dfrac{1 - cos A}{sin A}}}}$

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$\sf : \implies\:{\bold{\dfrac{1 }{sinA}- \dfrac{cos A}{sinA}}}$

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$\sf{\red{\boxed{\bold{\dfrac{1}{sinA} = cosec}}}}$

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$\sf{\red{\boxed{\bold{\dfrac{cosA}{sinA} = cot A}}}}$

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$\sf : \implies\:{\bold{ cosecA - cotA}}$

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$\sf{\underline{\underline{\large{\bold{RHS}}}}}$

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$\sf : \implies\:{\bold{ cosecA - cotA}}$

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$\sf{\underline{\underline{\large{\bold{Compare}}}}}$

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$\sf {\bold{ cosecA - cotA}}$ = $\sf {\bold{ cosecA - cotA}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.....Hence proved