Question

Prove that -

$\bf{ \frac{sin \: a - cos \: a + 1}{sin \: a + cos \: a - 1} = \frac{1}{sec \: a - tan \: a} }$

No irrelevant answers.

Step wise explanation needed.

Answer 1

L.H.S.

{(sina - cosa + 1)/ cosa } / {(sina + cosa - 1) / cosa}

= {tana - 1 + seca} / { tana + 1 - seca}

={tana + seca - 1} / {tana - seca + sec^2a - tan^2a}

= {tana + seca - 1} / (seca - tana) (tana + seca - 1)

= 1 / (seca - tana) = R.H.S.

since, sec^2a - tan^2a = 1

Hence, L.H.S.= R.H.S.

So, I hope it may help you mate.

Answer 2

Dividing by Cos Ø.

$\bf \implies{ \frac{ \frac{sin \: a}{cos \: a} - \frac{cos \: a }{cos \: a} + \frac{1}{cos \: a} }{ \frac{sin \: a}{cos \: a} + \frac{cos \: a}{cos \: a} - \frac{1}{cos \: a} } }$


$\bf{ \implies \: \frac{tan \: a - 1 + sec \: a}{tan \: a + 1 - sec \: a} }$



$\bf{ \implies \: \frac{tan \: a + sec \: a - 1}{tan \: a - sec \: a + 1} }$



$\bf{ \implies \frac{tan \: a \: + sec \: a - ( {sec}^{2}a - {tan}^{2}a) }{tan \: - sec \: a + 1} }$



$\bf{ \implies \frac{sa c \: a + tan \: a - (sec \: a + tan \: a)(sec \: a - tan \: a)}{tan \: a - sec \: a + 1} }$



$\bf{ \implies \frac{sec \: a + tan \: a(1 - (sec \: a - tan \: a))}{tan \: a - sec \: a + 1}}$



$\bf{ \implies\frac{ sec \: a + tan \: a( \cancel{1 - sec \: a + tan \: a})}{ \cancel{tan \: a - sec \: a + 1}}}$


$\bf{ \implies \: sec \: a + tan \: a}$


$\bf{ \implies \frac{sec \: a + tan \: a \: (sec \: a - tan \: a)}{sec \: a - tan \: a} }$


$\bf{ \implies \: \frac{{sec}^{2}a - {tan}^{2} a}{sec \: a - tan \: a}}$


$\bf{ \implies{ \frac{1}{sec \: a - tan \: a} }}$


As, sec^2 A - Tan^2 A =1