Question

prove that : sin A - cos A +1 / sinA +cos A- 1 = 1/ sec A - tan A​

Answer 1

Step-by-step explanation:

sin A - cos A +1 / sinA +cos A- 1 = 1/ sec A - tan A

here is your answer

Answer 2

Answer with Step-by-step explanation:

LHS

$\frac{\sin A-\cos A+1}{\sin A+\cos A-1}$

Divide numerator  and denominator by cos A

$\frac{\frac{sinA}{cosA}-\frac{cosA}{cosA}+\frac{1}{cos A}}{\frac{sinA}{cosA}+\frac{cosA}{cosA}-\frac{1}{cos A}}$

$\frac{tanA-1+secA}{tanA+1-secA}$

Using the formula

$tan A=\frac{sinA}{cos A}$

$sec A=\frac{1}{cos A}$

$\frac{tanA+secA-1}{tanA-secA+sec^2A-tan^2 A}$

Using the formula

$sec^2 A-tan^2A=1$

$\frac{tanA+secA-1}{tanA-secA+(secA-tanA)(secA+tanA)}$

$\frac{tanA+secA-1}{secA-tanA(-1+secA+tanA)}$

$\frac{1}{secA-tanA}$

LHS=RHS

Hence, proved.

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