Question

prove the following​

Answer 1

Answer:

here is your answer

Step-by-step explanation:

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Answer 2

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

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

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$\sf: \implies\: {\bold{ \dfrac{ SinA( 1 - CosA) + SinA( 1 + CosA)}{ ( 1 + CosA)( 1 - CosA)} = 2CosecA}}$

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

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

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

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$\sf: \implies\: {\bold{ \dfrac{ SinA - SinACosA + SinA + SinACosA}{ 1 - Cos^2A} = 2CosecA}}$

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$\sf: \implies\: {\bold{ \dfrac{ SinA - \cancel{SinACosA } + SinA + \cancel{SinACosA}}{ 1 - Cos^2A} = 2CosecA}}$

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$\sf: \implies\: {\bold{ \dfrac{ SinA + SinA }{ 1 - Cos^2A} = 2CosecA}}$

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

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$\sf: \implies\: {\bold{ \dfrac{ 2SinA}{ Sin^2A} = 2CosecA}}$

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$\sf: \implies\: {\bold{ \dfrac{ 2 \cancel{SinA }}{ SinA \times \cancel{SinA} } = 2CosecA}}$

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$\sf: \implies\: {\bold{ \dfrac{ 2 }{ SinA} = 2CosecA}}$

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

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$\sf: \implies\: {\bold{ 2 Cosec A = 2CosecA}}$

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⠀⠀⠀⠀⠀⠀⠀⠀.......Hence Proved

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