Question

Prove that: (sec A+ tan A- 1)/(tan A- sec A+1) = cos A/(1 - sin A)​

Answer 1

Step-by-step explanation:

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Answer 2

LHS

$:\implies \sf \dfrac{sec\ A+tan\ A-1}{tan\ A-sec\ A+1}$

• sec²A - tan²A = 1

$:\implies \sf \dfrac{sec\ A+tan\ A-(sec^2A-tan^2A)}{tan\ A-sec\ A+1}$

• a² - b² = (a+b)(a-b)

$\\ :\implies \sf \dfrac{(sec\ A+tan\ A)-[(sec\ A+tan\ A)(sec\ A-tan\ A)]}{tan\ A-sec\ A+1}$

$\\ :\implies \sf \dfrac{(sec\ A+tan\ A)(1-(sec\ A-tan\ A))}{1-sec\ A+tan\ A}$

$\\ :\implies \sf \dfrac{(sec\ A+tan\ A)(1-sec\ A+tan\ A)}{1-sec\ A+tan\ A}$

$\\ :\implies \sf \dfrac{(sec\ A+tan\ A)\cancel{(1-sec\ A+tan\ A)}}{\cancel{1-sec\ A+tan\ A}}$

$:\implies \sf sec\ A+tan\ A$

$:\implies \sf \dfrac{1}{cos\ A}+\dfrac{sin\ A}{cos\ A}$

$:\implies \sf \dfrac{1+sin\ A}{cos\ A}$

Rationalize the numerator

$:\implies \sf \dfrac{1+sin\ A}{cos\ A}\times \dfrac{1-sin\ A}{1-sin\ A}$

• (a+b)(a-b) = a² - b²

$:\implies \sf \dfrac{1-sin^2A}{cos\ A(1-sin\ A)}$

• ( 1 - sin²A ) = cos²A

$:\implies \sf \dfrac{cos^2A}{cos\ A(1-sin\ A)}$

$:\implies \sf \dfrac{cos\ A}{1-sin\ A}$

RHS