Question

Cos A - sin A +1/cosA +sin A -1 = cosec A + cot A prove

Answer 1

✨ QUESTION ✨

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

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✨IDENTITY USED✨

$\color{red} \boxed{\csc {}^{2} \alpha = > 1 + \cot {}^{2} \alpha }$

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L.H.S

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

divide both numerator and denominator by sin A

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

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

$\boxed{1 = > \csc ^{2} \alpha - \cot ^{2} \alpha }$

Use (a²-b²)=>{a-b}{a+b}

$a= > > \cot( \alpha ) \\ \\ b = > > \csc( \alpha )$

$\frac{ \cot( \alpha ) + \csc( \alpha ) - ( \csc( \alpha ) - \cot( \alpha ) \: )( \csc( \alpha ) + \cot( \alpha ) \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }$

$= > > \frac{ \cot( \alpha ) + \csc( \alpha )(1 - (\csc( \alpha ) - \cot( \alpha ) \: ) \: ) }{ \cot( \alpha ) + 1 - \csc( \alpha ) }$

$= > > \frac{ \cot( \alpha ) + \csc( \alpha ) - ( \: \: 1 - \csc( \alpha ) + \cot( \alpha ) \: \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }$

$= > > \frac{ \cot( \alpha ) + \csc( \alpha ) \: \: ( \: \: \cot( \alpha ) + 1 - \csc( \alpha ) \: \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }$

$= > > \color{blue} \boxed{\csc( \alpha ) + \cot( \alpha ) }$

Hence proved...

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Answer 2

Answer with Step-by-step explanation:

LHS

$\frac{cos A-sin A+1}{cos A+sin A-1}$

Divide numerator and denominator by sin A

$\frac{\frac{cos A-sin A+1}{sin A}}{\frac{cos A+sin A-1}{sin A}}$

$\frac{\frac{cos A}{sinA}-1+\frac{1}{sinA}}{\frac{cos A}{sinA}+1-\frac{1}{sinA}}$

We know that

$cot x=\frac{cos x}{sinx},cosec x=\frac{1}{sinx}$

By using the formula

$\frac{cot A-1+cosec A}{cot A+1-cosec A}$

$\frac{cot A+cosec A-1}{cot A+1- cosec A}$

We know that

$cosec^2 A-cot^2 A=1$

Using the identity

$\frac{(cot A+cosec A)-(cosec^2A-cot^2 A)}{cotA+1-cosec A}$

$\frac{(cosec A+cot A)-(cosec A+cot A)(cosecA-cot A)}{cot A+1-cosec A}$

Using identity:$a^2-b^2=(a+b)(a-b)$

$\frac{(cosecA+cot A)(1-cosec A+cot A)}{cot A+1-cosec A}$

$cot A+cosec A$=RHS

Hence, proved

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