Question

Evaluate:-
$\frac{ \cos( \beta ) - \sin( \beta ) + 1 }{ \cos( \beta ) \sin( \beta ) - 1} = co\sec( \beta ) + \cot( \beta )$
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Answer 1

Appropriate Question :- Prove that

$\rm \: \frac{ \cos\beta - \sin\beta + 1 }{ \cos \beta + \sin\beta - 1} = cosec \beta + cot \beta$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \frac{ \cos\beta - \sin\beta + 1 }{ \cos\beta + \sin\beta - 1}$

can be rewritten as

$\rm \: = \: \dfrac{\dfrac{cos\beta }{sin\beta } - \dfrac{sin\beta }{sin\beta } + \dfrac{1}{sin\beta } }{ \: \: \: \dfrac{cos\beta }{sin\beta } + \dfrac{sin\beta }{sin\beta } - \dfrac{1}{sin\beta } \: \: \: }$

$\rm \: = \: \dfrac{cot\beta - 1 + cosec\beta }{cot\beta + 1 - cosec\beta }$

can be re-arranged as

$\rm \: = \: \dfrac{(cot\beta + cosec\beta) - 1 }{cot\beta + 1 - cosec\beta }$

$\rm \: = \: \dfrac{(cot\beta + cosec\beta) - ( {cosec}^{2}\beta - {cot}^{2}\beta)}{cot\beta + 1 - cosec\beta }$

$\rm \: = \: \dfrac{(cot\beta + cosec\beta) - ( cosec\beta + cot\beta )(cosec\beta - cot\beta )}{cot\beta + 1 - cosec\beta }$

$\rm \: = \: \dfrac{(cosec\beta + cot\beta)\bigg[1 - (cosec\beta - cot\beta )\bigg]}{cot\beta + 1 - cosec\beta }$

$\rm \: = \: \dfrac{(cosec\beta + cot\beta)\bigg[1 - cosec\beta + cot\beta \bigg]}{cot\beta + 1 - cosec\beta }$

$\rm \: = \: cosec\beta + cot\beta$

Hence,

$\rm\implies \boxed{{\rm \: \frac{ \cos\beta - \sin\beta + 1 }{ \cos\beta + \sin\beta - 1} = cosec \beta + cot \beta \: }}$

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Formulae Used :-

$\boxed{ \rm{ \: \frac{1}{sinx} = cosecx \: }}$

$\boxed{ \rm{ \: \frac{cosx}{sinx} = cotx \: }}$

$\boxed{ \rm{ \: {cosec}^{2}x - {cot}^{2}x = 1 \: }}$

$\boxed{ \rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$