Question

prove that :-
I'll mark you as a brainliest

Answer 1

Solution.......

Let A = alpha

So,

$\frac{ \cot( \alpha ) - \cos( \alpha ) }{ \cot( \alpha ) + \cos( \alpha ) } = \frac{ \cosec( \alpha ) - 1}{ \cosec( \alpha ) + 1 }$

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☆LHS
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$\frac{ \cot( \alpha ) - \cos( \alpha ) }{ \cot( \alpha ) + \cos( \alpha ) } \\ \\ = \frac{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - \cos( \alpha ) }{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } + \cos( \alpha ) }$

$= \frac{ \cos( \alpha ) - \cos( \alpha ) \sin( \alpha ) }{ \cos( \alpha ) + \cos( \alpha ) \sin( \alpha ) } \\ \\ = \frac{ \cos \alpha (1 - \sin \alpha) }{ \cos \alpha (1 + \sin \alpha) } \\ \\ = \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) }$

Now,

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☆ RHS
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$\\ = \frac{ \cosec( \alpha ) - 1}{ \cosec( \alpha ) + 1}$
$\\ \\ = \frac{ \frac{1}{ \sin( \alpha ) } - 1}{ \frac{1}{ \sin( \alpha ) } + 1 }$

$\\ \\ = \frac{ \frac{1 - \sin( \alpha ) }{ \sin( \alpha ) } }{ \frac{1 + \sin( \alpha ) }{ \sin( \alpha ) } } \\ \\ \\ = \frac{1 - \sin( \alpha ) }{1 + \sin(alpha) }$

Therefore, LHS = RHS

HENCE, PROVED .

Answer 2

$\underline{\pink{\bold{TO \: \: PROVE}}}$ ➡

$\boxed{ \bold{\frac{cot \: A \: - cos \: A}{cot \: A + \: cos \: A} = \frac{cosec \: A- 1}{cosec \: A + 1} }}$

$\underline{\pink{\bold{PROOF}}}$ ➡

$\bold{L.H.S. = \frac{cot \: A - cos \: A}{cot \: A + cos \: A} } \\ \\ = \bold{ \frac{ \frac{cos \: A}{sin \: A} - cos \: A }{ \frac{cos \: A}{sin \: A} + cos \: A } } \\ \\ = \bold{ \frac{ \frac{cos \: A - cos \: A.sin \: A}{sin \: A} }{ \frac{cos \: A + cos \: A.sin \: A}{sin \: A} } } \\ \\ = \bold{ \frac{cos \: A -cos \: A .sin \: A }{cos \: A + cos \: A.sin \: A} } \\ \\ = \bold{ \frac{cos \: A \: (1 - sin \: A)}{cos \: A \: (1 + sin \: A)} } \\ \\ = \bold{ \frac{1 -sin \: A }{1 + sin \: A} } \\ \\ = \bold{ \frac{ \frac{1 - sin \: A}{sin \: S} }{ \frac{1 + sin \: A}{sin \: A} } } \\ \\ = \bold{ \frac{ \frac{1}{sin \: A} - \frac{sin \: a}{sin \: a} }{ \frac{1}{sin \: A} + \frac{sin \: A}{sin \: A} } } \\ \\ = \bold{ \frac{cosec \: A - 1}{cosec \: A + 1} } \\ \\ = \bold{R.H.S.}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Hence \: \: Proved.}$

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✝$\bold{ \red{ \mathfrak{THANKS...}}}$ ✝