Question

For best users and moderators :D
Prove that -
$\dfrac{ \cos( \beta ) - \sin( \beta ) + 1 }{ \cos( \beta ) + \sin( \beta ) - 1} = \cosec( \beta ) + \cot( \beta )$
​

Answer 1

$\\ \color{maroon}=\dfrac{\frac{\cos \beta }{\sin \beta }- \cancel\frac{\sin \beta }{\sin \beta }+\frac{1}{\sin \beta }}{\frac{\cos \beta }{\sin \beta }+ \cancel\frac{\sin \beta }{\sin \beta }-\frac{1}{\sin \beta }}$

$\\ \\ \color{blue}=\frac{\cot \beta -1+\operatorname{cosec} \beta }{\cot \beta +1-\operatorname{cosec} \beta }$

$\\ \\ \red{ =\frac{\cot \beta +\operatorname{cosec} \beta -1}{1+\cot \beta -\operatorname{cosec} \beta } }$

$\color{magenta} \begin{array}{l}\\ \\ \\ \\{=\dfrac{\cot \beta +\operatorname{cosec} \beta +\cot ^{2} \beta -\operatorname{cosec}^{2} \beta }{1+\cot \beta -\operatorname{cosec} \beta }} \\ \\ \\ \\ = \dfrac{(\cot \beta -\operatorname{cosec} \beta )(\cot \beta -\operatorname{cosec} \beta )(\cot \beta -\operatorname{cosec} \beta )}{1+\cos \beta +\operatorname{cosec} \beta } \\ \\ \\ \\ =\dfrac{\cot \beta +\operatorname{cosec} \beta \cancel{ (1+\cot \beta -\operatorname{cosec} \beta )}} {\cancel{1+\cot \beta -\operatorname{cosec} \beta }} \\ \\ \\ \\ =\cot \beta +\operatorname{cosec} \beta \end{array}$

Answer 2

To Find:

Prove:

$\dfrac{ \cos( \beta ) - \sin( \beta ) + 1 }{ \cos( \beta ) + \sin( \beta ) - 1} = \cosec( \beta ) + \cot( \beta )$

$\begin{gathered} \\ \color{violet}=\dfrac{\frac{\cos \beta }{\sin \beta }- \cancel\frac{\sin \beta }{\sin \beta }+\frac{1}{\sin \beta }}{\frac{\cos \beta }{\sin \beta }+ \cancel\frac{\sin \beta }{\sin \beta }-\frac{1}{\sin \beta }}\end{gathered} =$

$\begin{gathered} \\ \\ \tt\color{red}=\frac{\cot \beta -1+\operatorname{cosec} \beta }{\cot \beta +1-\operatorname{cosec} \beta } \end{gathered}$

$\begin{gathered} \\ \\ \red{ =\frac{\cot \beta +\operatorname{cosec} \beta -1}{1+\cot \beta -\operatorname{cosec} \beta } }\end{gathered}$

$\begin{gathered} \color{cyan} \begin{array}{l}\\ \\ \\ \\{=\dfrac{\cot \beta +\operatorname{cosec} \beta +\cot ^{2} \beta -\operatorname{cosec}^{2} \beta }{1+\cot \beta -\operatorname{cosec} \beta }} \\ \\ \\ \\ = \dfrac{(\cot \beta -\operatorname{cosec} \beta )(\cot \beta -\operatorname{cosec} \beta )(\cot \beta -\operatorname{cosec} \beta )}{1+\cos \beta +\operatorname{cosec} \beta } \\ \\ \\ \\ =\dfrac{\cot \beta +\operatorname{cosec} \beta \cancel{ (1+\cot \beta -\operatorname{cosec} \beta )}} {\cancel{1+\cot \beta -\operatorname{cosec} \beta }} \\ \\ \\ \\ =\cot \beta +\operatorname{cosec} \beta \end{array}\end{gathered}$