Math, asked by estherjoseph3872, 1 year ago

(CosecA-cotA)square=1-cosA/1+cosa

Answers

Answered by Anonymous
111

AnswEr :

To Prove :

 \rm( \csc(A)  -  \cot(A) )^{2}  =  \dfrac{1 -  \cos(A) }{1 +  \cos(A) }

Proof :

I'm taking RHS to Prove this Identity -

\longrightarrow \sf\dfrac{1 -  \cos(A) }{1 +  \cos(A) }

\longrightarrow \sf\dfrac{1 -  \cos(A) }{1 +  \cos(A) }   \times 1

⠀⠀⠀⋆ Multiplying by 1 - cos(A) / 1 - cos(A)

\longrightarrow \sf\dfrac{1 -  \cos(A) }{1 +  \cos(A) }  \times \dfrac{1 -  \cos(A) }{1  -   \cos(A) }

\longrightarrow \sf\dfrac{(1 -  \cos(A) )^{2} }{1  -   \cos^{2} (A) }

⠀⠀⠀⋆ (a - b)² = a² + b² - 2ab

⠀⠀⠀⋆ (1 - cos²(A)) = sin²(A)

\longrightarrow \sf\dfrac{1  +  \cos ^{2} (A)  - 2 \cos(A) }{ \sin^{2} (A) }

\longrightarrow \sf\dfrac{1}{ \sin ^{2} (A) } + \dfrac{ \cos^{2} (A) }{ \sin ^{2} (A) } -  \dfrac{2 \cos(A) }{ \sin ^{2} (A) }

⠀⠀⠀⋆ 1 / sin²(A) = cosec²(A)

⠀⠀⠀⋆ cos²(A) / sin²(A) = cot²(A)

⠀⠀⠀⋆ cos(A) / sin²(A) = cot(A)cosec(A)

\longrightarrow \sf \csc^{2} (A)  +  \cot^{2} (A) - 2 \cot(A)   \csc(A)

⠀⠀⠀⋆ (a² + b² + 2ab) = (a - b)²

\longrightarrow \large \sf( \csc(A)  -  \cot(A) )^{2}

 \therefore \boxed{ \rm( \csc(A)  -  \cot(A) )^{2}  =  \dfrac{1 -  \cos(A) }{1 +  \cos(A) } }

Answered by kaushik05
20

 \huge \red{ \mathfrak{solution}}

To prove :

 {( \csc( \alpha )  -  \cot( \alpha )) }^{2}  =  \frac{1 -  \cos( \alpha ) }{1 +  \cos( \alpha ) }  \\  \\

LHS

 \leadsto \: ( { \csc( \alpha )  -  \cot( \alpha )) }^{2}  \\  \\  \leadsto \:( \frac{1}{ \sin( \alpha ) }  -  \frac{ \cos( \alpha ) }{ \sin( \alpha ) } ) ^{2}  \\  \\  \leadsto( \frac{1 -  \cos( \alpha ) }{ \sin( \alpha ) } )^{2}  \\  \\   \leadsto \:  \frac{(1 -  \cos( \alpha )) ^{2}  }{ { \sin }^{2}  \alpha }  \\  \\  \leadsto \:   \frac{( {1 -  \cos( \alpha ) })^{2} }{1 -  \cos ^{2} ( \alpha ) }  \\  \\  \leadsto  \:  \frac{(1 -  \cos( \alpha ) )(1 -  \cos( \alpha ) )}{(1 -  \cos( \alpha ))(1 +  \cos( \alpha ))  }  \\  \\  \leadsto \:  \cancel  \frac{1 -  \cos( \alpha ) }{1 -  \cos( \alpha ) }  \times  \frac{1 -  \cos( \alpha ) }{1 +  \cos( \alpha ) }  \\  \\ \leadsto \:  \frac{1 -  \cos( \alpha ) }{1 +  \cos( \alpha ) }

LHS = RHS

 \boxed{  \green{\bold{  \mathfrak{proved}}}}

Formula :

a^2-b^2=(a+b)(a-b)

csc@= 1/sin@

cot@= cos@/sin@

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