[tex]\rule{200pt}{4pt}[/tex]
[tex]{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}} = {\dfrac{cos{\theta}}{1 - sin{\theta}}}[/tex]
[tex]\\\\[/tex]
[tex]{\rightarrow{\sf{\dfrac{\dfrac{1}{cos{\theta}} + {\dfrac{sin {\theta}}{cos{\theta}} - 1}}{{\dfrac{sin{\theta}}{\dfrac{1}{cos{\theta}} } + 1}}}}}[/tex]
[tex]\sf{\rightarrow{\dfrac{\dfrac{ 1 + Sin {\theta} - cos {\theta}}{cos{\theta}}}{{\dfrac{Sin {\theta} - 1 + cos {\theta}}{cos{\theta}}}}}}[/tex]
[tex]\sf{\rightarrow{\dfrac{1 + sin {\theta} - cos{\theta}}{sin {\theta} - 1 + cos{\theta}}}}[/tex]
[tex]\\[/tex]
By using : [tex]\sf{cos²{\theta} + sin²{\theta} = 1}[/tex]
[tex]\sf{\rightarrow{\dfrac{cos² {\theta} - cos{\theta}}{sin {\theta} - 1 + sin²{\theta}}}}[/tex]
[tex]\sf{\rightarrow{\dfrac{cos{\theta} }{- 1 + sin{\theta}}}}[/tex]
[tex]{\sf{\rightarrow{\dfrac{cos{\theta}}{1 - sin{\theta}}}}}[/tex]
[tex]{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}}[/tex]= [tex]{\dfrac{cos{\theta}}{1 - sin{\theta}}}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \large{\sf{ \implies{Hence\:proved}}}[/tex]
\rule{200pt}{4pt}
{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}} = {\dfrac{cos{\theta}}{1 - sin{\theta}}}→
tanθ−secθ+1
secθ+tanθ−1
=
1−sinθ
cosθ
\begin{gathered}\\\\\end{gathered}
{\rightarrow{\sf{\dfrac{\dfrac{1}{cos{\theta}} + {\dfrac{sin {\theta}}{cos{\theta}} - 1}}{{\dfrac{sin{\theta}}{\dfrac{1}{cos{\theta}} } + 1}}}}}→
1
sinθ
+1
+
−1
\sf{\rightarrow{\dfrac{\dfrac{ 1 + Sin {\theta} - cos {\theta}}{cos{\theta}}}{{\dfrac{Sin {\theta} - 1 + cos {\theta}}{cos{\theta}}}}}}→
Sinθ−1+cosθ
1+Sinθ−cosθ
\sf{\rightarrow{\dfrac{1 + sin {\theta} - cos{\theta}}{sin {\theta} - 1 + cos{\theta}}}}→
sinθ−1+cosθ
1+sinθ−cosθ
\begin{gathered}\\\end{gathered}
By using : \sf{cos²{\theta} + sin²{\theta} = 1}cos
2
θ+sin
θ=1
\sf{\rightarrow{\dfrac{cos² {\theta} - cos{\theta}}{sin {\theta} - 1 + sin²{\theta}}}}→
sinθ−1+sin
θ
cos
θ−cosθ
\sf{\rightarrow{\dfrac{cos{\theta} }{- 1 + sin{\theta}}}}→
−1+sinθ
{\sf{\rightarrow{\dfrac{cos{\theta}}{1 - sin{\theta}}}}}→
{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}}→
= {\dfrac{cos{\theta}}{1 - sin{\theta}}}
\: \: \: \: \: \: \: \: \: \: \: \: \: \large{\sf{ \implies{Hence\:proved}}}⟹Henceproved
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Answers & Comments
[tex]\rule{200pt}{4pt}[/tex]
[tex]{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}} = {\dfrac{cos{\theta}}{1 - sin{\theta}}}[/tex]
[tex]\\\\[/tex]
[tex]{\rightarrow{\sf{\dfrac{\dfrac{1}{cos{\theta}} + {\dfrac{sin {\theta}}{cos{\theta}} - 1}}{{\dfrac{sin{\theta}}{\dfrac{1}{cos{\theta}} } + 1}}}}}[/tex]
[tex]\\\\[/tex]
[tex]\sf{\rightarrow{\dfrac{\dfrac{ 1 + Sin {\theta} - cos {\theta}}{cos{\theta}}}{{\dfrac{Sin {\theta} - 1 + cos {\theta}}{cos{\theta}}}}}}[/tex]
[tex]\\\\[/tex]
[tex]\sf{\rightarrow{\dfrac{1 + sin {\theta} - cos{\theta}}{sin {\theta} - 1 + cos{\theta}}}}[/tex]
[tex]\\[/tex]
By using : [tex]\sf{cos²{\theta} + sin²{\theta} = 1}[/tex]
[tex]\\\\[/tex]
[tex]\sf{\rightarrow{\dfrac{cos² {\theta} - cos{\theta}}{sin {\theta} - 1 + sin²{\theta}}}}[/tex]
[tex]\\\\[/tex]
[tex]\sf{\rightarrow{\dfrac{cos{\theta} }{- 1 + sin{\theta}}}}[/tex]
[tex]\\\\[/tex]
[tex]{\sf{\rightarrow{\dfrac{cos{\theta}}{1 - sin{\theta}}}}}[/tex]
[tex]\\\\[/tex]
[tex]{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}}[/tex]= [tex]{\dfrac{cos{\theta}}{1 - sin{\theta}}}[/tex]
[tex]\\\\[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \large{\sf{ \implies{Hence\:proved}}}[/tex]
[tex]\rule{200pt}{4pt}[/tex]
\rule{200pt}{4pt}
{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}} = {\dfrac{cos{\theta}}{1 - sin{\theta}}}→
tanθ−secθ+1
secθ+tanθ−1
=
1−sinθ
cosθ
\begin{gathered}\\\\\end{gathered}
{\rightarrow{\sf{\dfrac{\dfrac{1}{cos{\theta}} + {\dfrac{sin {\theta}}{cos{\theta}} - 1}}{{\dfrac{sin{\theta}}{\dfrac{1}{cos{\theta}} } + 1}}}}}→
cosθ
1
sinθ
+1
cosθ
1
+
cosθ
sinθ
−1
\begin{gathered}\\\\\end{gathered}
\sf{\rightarrow{\dfrac{\dfrac{ 1 + Sin {\theta} - cos {\theta}}{cos{\theta}}}{{\dfrac{Sin {\theta} - 1 + cos {\theta}}{cos{\theta}}}}}}→
cosθ
Sinθ−1+cosθ
cosθ
1+Sinθ−cosθ
\begin{gathered}\\\\\end{gathered}
\sf{\rightarrow{\dfrac{1 + sin {\theta} - cos{\theta}}{sin {\theta} - 1 + cos{\theta}}}}→
sinθ−1+cosθ
1+sinθ−cosθ
\begin{gathered}\\\end{gathered}
By using : \sf{cos²{\theta} + sin²{\theta} = 1}cos
2
θ+sin
2
θ=1
\begin{gathered}\\\\\end{gathered}
\sf{\rightarrow{\dfrac{cos² {\theta} - cos{\theta}}{sin {\theta} - 1 + sin²{\theta}}}}→
sinθ−1+sin
2
θ
cos
2
θ−cosθ
\begin{gathered}\\\\\end{gathered}
\sf{\rightarrow{\dfrac{cos{\theta} }{- 1 + sin{\theta}}}}→
−1+sinθ
cosθ
\begin{gathered}\\\\\end{gathered}
{\sf{\rightarrow{\dfrac{cos{\theta}}{1 - sin{\theta}}}}}→
1−sinθ
cosθ
\begin{gathered}\\\\\end{gathered}
{\rightarrow{\sf{\dfrac{sec {\theta} + tan {\theta} - 1}{tan{\theta} - sec{\theta} + 1}}}}→
tanθ−secθ+1
secθ+tanθ−1
= {\dfrac{cos{\theta}}{1 - sin{\theta}}}
1−sinθ
cosθ
\begin{gathered}\\\\\end{gathered}
\: \: \: \: \: \: \: \: \: \: \: \: \: \large{\sf{ \implies{Hence\:proved}}}⟹Henceproved
\rule{200pt}{4pt}