Question

prove that(1-cosA/1+cosA)=( cosecA-cotA)
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Answer 1

Answer:

Simplifying LHS

Simplifying LHS1-cosA/1+cosA

(1-cosA)/(1+cosA)*(1-cosA)/(1-cosA)

(1-cosA) ^2/1-cos^2A

(1-cosA) ^2/sin^2A

[1-cosA/sinA]^2

[1/sinA - cosA/sinA]^2

[cosecA-cotA]^2

HENCE PROVED

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Answer 2

$\sf{\red{\boxed{\bold{Solution}}}}$

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$\sf :\implies\:{\dfrac{1 - CosA}{1+cosA} = ( cosecA - CotA)^2 }$

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• Rationalize the denominator

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$\sf :\implies\:{\dfrac{1 - CosA}{1+cosA} \times \dfrac{ 1 - CosA}{ 1 - CosA} = ( cosecA - CotA)^2 }$

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$\sf :\implies\:{\dfrac{(1 - CosA)^2}{(1+cosA)( 1 - CosA) } = ( cosecA - CotA)^2 }$

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

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$\sf :\implies\:{\dfrac{( 1 - CosA)^2}{1^2 - cos^2A} = ( cosecA - CotA)^2 }$

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$\sf{\red{\boxed{\bold{( 1 - Cos^2A) = Sin^2A}}}}$

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$\sf :\implies\:{\dfrac{(1 - CosA)^2}{Sin^2A} = ( cosecA - CotA)^2 }$

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$\sf :\implies\:{\bigg\lgroup \dfrac{1 - CosA}{SinA}\bigg\rgroup ^2 = ( cosecA - CotA)^2 }$

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$\sf :\implies\:{\bigg\lgroup \dfrac{1}{SinA} - \dfrac{CosA}{SinA}\bigg\rgroup ^2 = ( cosecA - CotA)^2 }$

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$\sf{\red{\boxed{\bold{\dfrac{1}{SinA} = CosecA}}}}$

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$\sf{\red{\boxed{\bold{\dfrac{CosA}{SinA} = Cot A}}}}$

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$\sf :\implies\:{\big\lgroup Cosec A - CotA \big\rgroup ^2 = ( cosecA - CotA)^2 }$

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LHS = RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.........Hence Proved

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