Question

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Answer 1

$\huge\underline\mathfrak\red{Question:-}$

$\tt{To\: prove:-}$$\tt \frac{ \sin θ- \cos θ+ 1}{ \sin θ + \cos θ- 1 } = \frac{1}{ \secθ - \tan θ}$

$\huge\underline\mathfrak\green{Answer:-}$

In this problem we will use suitable formulas like,

${ \tan}^{2}θ - { \sec }^{2} θ= - 1$

L.H.S

$\rightarrow\frac{ \sin θ - \cosθ+ 1 }{ \sin θ + \cosθ - 1 }$

[Dividing by cosθ on both numerator and denominator]

$\rightarrow\tt \frac{ \frac{ \sin θ - \cos θ + 1 }{ \cos θ } }{ \frac{ \sin θ+ \cosθ- 1}{ \cos θ } }$

$\rightarrow\tt \frac{ \tan θ - 1 + \sec θ }{ \tan θ + 1 - \sec θ }$

$\rightarrow\tt \frac{( \tan θ + \sec θ ) - 1}{( \tan θ - \sec θ + 1 }$

$\rightarrow\tt \frac {( {\tan}^{2}θ + \secθ- 1)( \tan θ - \sec θ ) }{[( \tan θ - \sec θ ) +1]( \tanθ - \sec θ )}$

$\rightarrow\tt \frac{ ({ \tan }^{2} θ - { \sec }^{2} θ ) - ( \tan θ - \sec θ ) }{( \tan θ- \sec θ + 1)( \tan θ - \sec θ ) }$

$\rightarrow\tt \frac{ - 1 - \tan θ + \sec θ }{( \tan θ - \sec θ + 1)( \tan θ- \sec θ )}$

$\rightarrow\tt \frac{ - 1 - ( \tan θ - \sec θ ) }{( \tan θ - \sec θ + 1)( \tan θ- \secθ ) }$

$\rightarrow\tt \frac{ - 1}{ \tan θ - \sec θ }$

$\rightarrow\tt \frac{ 1}{ \sec θ - \tan θ}$

= R.H.S

Answer 2

Answer:

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