Math, asked by beenaaditya2015, 4 months ago

Prove that : 1-cosA/1+cosA =( cotA - cosecA) square​

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Answered by BrainlyRish
7

Given : \sf{\dfrac{1 - \cos A }{1 + \cos A } = ( \cot A - \cosec A )^{2} }\\

To Prove: \sf{\dfrac{1 - \cos A }{1 + \cos A } = ( \cot A - \cosec A )^{2} }\\

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⠀⠀⠀⠀\star\sf{\dfrac{1 - \cos A }{1 + \cos A } = ( \cot A - \cosec A )^{2} }\\

Here ,

⠀⠀⠀⠀\implies \sf{L.H.S \: = \dfrac{1 - \cos A }{1 + \cos A }  }\\

⠀⠀⠀⠀\implies \sf{R.H.S = ( \cot A - \cosec A )^{2} }\\

⠀⠀⠀⠀⠀⠀\underline {\frak{\star\:Now \: By \: Solving \: the \: L.H.S \:  \::}}\\

⠀⠀⠀⠀:\implies \sf{L.H.S \: = \dfrac{1 - \cos A }{1 + \cos A }  }\\

⠀⠀⠀⠀:\implies \sf{ \dfrac{1 - \cos A }{1 + \cos A }  }\\

▪︎ Multiplying both and numerator and denominator by ( 1 - cos A ) .

⠀⠀⠀⠀:\implies \sf{ \dfrac{1 - \cos A }{1 + \cos A } \times \dfrac{1 - \cos A }{1 - \cos A }  }\\

As We know that ,

  • (a+b)(a-b) = (a - b)²

⠀⠀⠀⠀:\implies \sf{ \dfrac{1 - \cos A }{1 + \cos A } \times \dfrac{1 - \cos A }{1 - \cos A }  }\\

⠀⠀⠀⠀:\implies \sf{ \dfrac{(1 - \cos A )^{2}}{1 - \cos^{2} A }   }\\

As , We know that ,

  • \star\sin^{2}\theta = 1 - \cos^{2} A

⠀⠀⠀⠀:\implies \sf{ \dfrac{(1 - \cos A )^{2}}{ \sin^{2} A }   }\\

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

Or ,

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

As, We know that,

  • \star\dfrac{1}{\sin A} = \cosec A

⠀⠀⠀⠀:\implies \sf{\bigg( \cosec A - \dfrac{\cos A }{\sin A} \bigg) ^{2}   }\\

As , We know that ,

  • \star\dfrac{\cos A }{\sin A} = \cot A

⠀⠀⠀⠀:\implies \sf{\bigg( \cosec A - \cot \bigg) ^{2}   }\\

Or,

⠀⠀⠀⠀:\implies \sf{L.H.S\:= \bigg( \cot A - \cosec \bigg) ^{2}   }\\

Therefore,

⠀⠀⠀⠀:\implies \sf{L.H.S\:= \bigg( \cot A - \cosec \bigg) ^{2}   }\\

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀\implies \sf{L.H.S\:= \bigg( \cot A - \cosec \bigg) ^{2}   }\\

⠀⠀⠀⠀\implies \sf{R.H.S = ( \cot A - \cosec A )^{2} }\\

Therefore,

⠀⠀⠀⠀⠀\implies \sf{L.H.S = R.H.S }\\

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\therefore \sf{\bf{Hence ,\:Proved .} }\\

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