Question

√(1-cos a/1+ cos a) = sin a/ 1+cos a​

Answer 1

Answer:

Step-by-step explanation:

LHS:

sqr(1-cosa/1+cosa)

multiply numerator and denominator with 1+cosa

=sqrt((1-cosa)*(1+cosa)/(1-cos^2a))

=sqrt(1-cos^2a)/(1+cosa)^2))

=sqrt(sin^2a/((1+cosa)^2)

=sina/1+cosa

=RHS

Answer 2

$\sf{\underline{\boxed{\purple{\large{\bold{ Solution }}}}}}$

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$\sf :\implies\:{\bold{ \sqrt{ \dfrac{ 1 - CosA}{ 1 + Cos A}} = \dfrac{SinA}{1 + CosA} }}$

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$\sf :\implies\:{\bold{ \sqrt{ \dfrac{ 1 - CosA}{ 1 + Cos A} \times \dfrac{1 + cosA}{1 + CosA}} = \dfrac{SinA}{1 + CosA} }}$

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

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

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

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$\sf :\implies\:{\bold{ \sqrt{ \dfrac{ Sin^2A}{( 1 + Cos A)^2}} = \dfrac{SinA}{1 + CosA} }}$

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

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$\sf :\implies\:{\bold{ \sqrt{ \dfrac{ SinA}{ 1 + Cos A}} = \dfrac{SinA}{1 + CosA} }}$

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LHS = RHS

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀......Hence Proved