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${\sf{{\dfrac{sinA\: -\: cosA\: +\: 1}{sinA\: +\: cosA\: -\: 1}}\: =\: {\dfrac{cosA}{1\: -\: sinA}}}}$
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Answer 1

Answer:

Let us first divide both the numerator and the denominator with Cos A. Hence we get:

$\implies \dfrac{tan\:A - 1 + sec\:A}{tan\:A + 1 - sec\:A}\\\\\\\text{Rearranging the terms we get:}\\\\\\\implies \dfrac{tan\:A + sec\:A - 1}{tan\:A - sec\:A + 1}\\\\\\\text{Substituting 1 = } (sec^2\:A - tan^2A) \: \text{we get:}\\\\\\\implies \dfrac{tan\:A + sec\:A - (sec^2A - tan^2A)}{tan\:A - sec\:A + 1}\\\\\\\text{Simplifying the identity } a^2-b^2 = (a+b)(a-b) \text{ we get:}\\\\\\\implies \dfrac{tan\:A + sec\:A - (sec\:A+tan\:A)(sec\:A-tan\:A)}{tan\:A - sec\:A + 1}$

Taking (Tan A + Sec A) commonly outside, in the numerator we get:

$\implies \dfrac{(tan\:A + sec\:A)(tan\:A - sec\:A+1)}{tan\:A - sec\:A + 1}\\\\\\\implies \boxed{ tan\:A + sec\:A}\\\\\text{Multiplying by sec A - tan A to the numerator and denominator we get:}\\\\\\\implies \dfrac{tan\:A + sec\:A}{1} \times \dfrac{sec\:A - tan\:A}{sec\:A - tan\:A}\\\\\\\implies \dfrac{sec^2A - tan^2A}{sec\:A - tan\:A}\\\\\implies \dfrac{ 1}{sec\:A - tan\:A}\\\\\\\text{Multiplying by cos\:A to both the Nr and Dr, we get:}\\\\\implies \boxed{ \bf{ \dfrac{cos\:A}{1 - sin\:A}}}$

Hence Proved.

Answer 2

${\boxed{\underline{ \orange{\tt{Required \: \: answer:-}}}}}$

★QUESTION:-

${\sf{{\dfrac{sinA\: -\: cosA\: +\: 1}{sinA\: +\: cosA\: -\: 1}}\: =\: {\dfrac{cosA}{1\: -\: sinA}}}}$

★EXPLANATION:-

We will solve LHS and RHS

${\boxed{\underline{\bf{\red{L .H.S}}}}}$

Divide Both the sides by Cos A in each term

$: \implies \sf \large{ \frac{ \frac{SinA }{CosA} - \frac{CosA}{CosA} + \frac{1}{CosA} }{ \frac{SinA}{CosA} + \frac{CosA}{CosA} - \frac{1}{CosA} } }$

$: \implies \sf \large{ \frac{tan A + secA - 1}{tan A -secA + 1 } }$

$: \implies \sf \large{ \frac{(secA + tan A ) - ( {secA}^{2} - {tan A}^{2} }{tan A - secA + 1} }$

$: \implies \sf \large{ \frac{(secA + tan A) - (secA - tan A)(secA + tan A)}{tan A -secA + 1 } }$

$: \implies \sf \large{ \frac{(secA + tan A)( \cancel{1 - secA + tan A)}}{ \cancel{1 - secA + tan A}} }$

$: \implies \sf \large \purple{secA + tan A}$

${\boxed{\underline{\bf{\red{R .H.S}}}}}$

$\sf{\dfrac {cosA}{1\: -\: sinA}}$

Reverse The denominator and Numerator with the reverse of - sign also.

$: \implies \sf \large{ \frac{1 + SinA }{CosA} }$

Take the fraction with both of them

$: \implies \sf \large{ \frac{1}{CosA} + \frac{SinA}{CosA} }$

$: \implies \sf \large \purple{secA + tan A}$

So,

LHS = RHS

HENCE PROVED.