Question

$prove \: that \: \frac{1 + \sin( \alpha ) - \cos( \alpha ) }{ \cos( \alpha ) - 1 + \sin( \alpha ) } - \frac{1 - \sin( \alpha ) + \cos( \alpha ) }{ \cos( \alpha ) + 1 + \sin( \alpha ) } = 2 \tan( \alpha )$
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Answer 1

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

Consider,

$\purple{\rm :\longmapsto\:\dfrac{1 + sin \alpha - cos\alpha }{cos\alpha - 1 + sin\alpha } }$

can be rewritten as

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

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

$\purple{\rm \: = \: \dfrac{sec\alpha + tan\alpha - 1}{1 - sec\alpha + tan\alpha } }$

$\purple{\rm \: = \: \dfrac{sec\alpha + tan\alpha - ( {sec}^{2} \alpha - {tan}^{2} \alpha )}{1 - sec\alpha + tan\alpha } }$

$\purple{\rm \: = \: \dfrac{sec\alpha + tan\alpha - (sec\alpha + tan\alpha)(sec\alpha - tan\alpha )}{1 - sec\alpha + tan\alpha } }$

$\purple{\rm \: = \: \dfrac{(sec\alpha + tan\alpha)(1 - sec\alpha + tan\alpha )}{1 - sec\alpha + tan\alpha } }$

$\purple{\rm \: = \: sec\alpha + tan\alpha }$

Thus,

$\purple{\rm \implies\:\boxed{ \tt{ \: \dfrac{1 + sin \alpha - cos\alpha }{cos\alpha - 1 + sin\alpha } = sec\alpha + tan\alpha }}}$

Now Consider :-

$\red{\rm :\longmapsto\:\dfrac{1 - sin\alpha + cos\alpha }{cos\alpha + 1 + sin\alpha } }$

can be rewritten as

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

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

$\red{\rm \: = \: \dfrac{sec\alpha - tan\alpha + 1}{1 + sec\alpha + tan\alpha } }$

$\red{\rm \: = \: \dfrac{sec\alpha - tan\alpha + ( {sec}^{2}\alpha - {tan}^{2}\alpha )}{1 + sec\alpha + tan\alpha } }$

$\red{\rm \: = \: \dfrac{sec\alpha - tan\alpha + (sec\alpha + tan\alpha)(sec\alpha - tan\alpha )}{1 + sec\alpha + tan\alpha } }$

$\red{\rm \: = \: \dfrac{(sec\alpha - tan\alpha)(1 + sec\alpha + tan\alpha )}{1 + sec\alpha + tan\alpha } }$

$\red{\rm \: = \: sec\alpha - tan\alpha }$

Thus,

$\red{\rm \implies\:\boxed{ \tt{ \: \dfrac{1 - sin\alpha + cos\alpha }{cos\alpha + 1 + sin\alpha } = sec\alpha - tan\alpha }}}$

Now, Consider

$\blue{\rm :\longmapsto\:\dfrac{1 + sin \alpha - cos\alpha }{cos\alpha - 1 + sin\alpha } \: - \: \dfrac{1 - sin\alpha + cos\alpha }{cos\alpha + 1 + sin\alpha } }$

On substituting the values of both, evaluated above, we get

$\blue{\rm \: = \: (sec\alpha + tan\alpha ) - (sec\alpha - tan\alpha )}$

$\blue{\rm \: = \: sec\alpha + tan\alpha - sec\alpha + tan\alpha }$

$\blue{\rm \: = \: 2tan\alpha }$

Hence,

$\blue{\boxed{ \tt{ \: \dfrac{1 + sin \alpha - cos\alpha }{cos\alpha - 1 + sin\alpha } \: - \: \dfrac{1 - sin\alpha + cos\alpha }{cos\alpha + 1 + sin\alpha } = 2tan\alpha}}}$

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Formula Used :-

$\boxed{ \tt{ \: \frac{1}{cosx} = secx}}$

$\boxed{ \tt{ \: \frac{sinx}{cosx} = tanx}}$

$\boxed{ \tt{ \: {sec}^{2}x - {tan}^{2}x = 1 \: }}$