Question

Prove that (Sin A-Cos A+1) / (Sin A+Cos A-1) =1/ (Sec A-tan A)​

Answer 1

Answer:

Proved.

Step-by-step explanation:

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

= $\frac{(sin A - cos A + 1)(sin A + cos A + 1)}{(sin A + cos A - 1)(sin A + cos A + 1)}$

= $\frac{(sin A + 1)^{2} - cos^{2} A}{(sin A + cos A)^{2} - 1^{2}}$

= $\frac{sin^{2}A + 2sin A + 1 - cos^{2}A}{sin^{2}A + cos^{2}A + 2sin AcosA - 1}$

= $\frac{sin^{2}A + 2sinA + sin^{2}A}{1 + 2sinAcosA - 1}$

= $\frac{2sin^{2}A + 2sinA}{2sinAcosA}$

= $\frac{2sinA(sin A + 1)}{2sinAcosA}$

= $\frac{sin A + 1}{cosA}$

= $\frac{sin A}{cos A} + \frac{1}{cosA}$

= $tan A + sec A$

= $\frac{(sec A + tan A)(sec A - tan A)}{(sec A - tan A)}$

= $\frac{sec^{2}A - tan^{2}A}{sec A - tan A}$

= $\frac{1}{sec A - tan A}$

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