Question

Prove that :
sinA - cosA + 1 / sinA + cosA - 1 = 1 / secA - tanA

Answer 1

Answer:

L.H.S=R.H.S

Step-by-step explanation:

We have been given that

\frac{sinA - cosA + 1}{sinA + cosA - 1} = \frac{1}{secA - tanA}

Let, L.H.S

\frac{sinA - cosA + 1}{sinA + cosA - 1}

\frac{\frac{sinA}{cosA} - \frac{cosA}{cosA} + \frac{1}{cosA}}{\frac{sinA}{cosA} + \frac{cosA}{cosA} - \frac{1}{cosA}}

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

Formula, sec^{2}A - tan^{2}A = 1

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

\frac{tanA + secA - [(secA - tanA)(secA + tanA)]}{tanA + 1 - secA}

\frac{tanA + secA [1 - (secA - tanA)]}{tanA + 1 - secA}

[1 - (secA - tanA)] cancel out

Then,  {tanA + secA

Multiple and divide by secA -   tanA

Then,  \frac{(tanA + secA)\times ( secA -   tanA )}{ secA -   tanA }

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

Using this formula, sec^{2}A - tan^{2}A = 1

\frac{1}{ secA -   tanA }

R.H.S

Answer 2

Hence , the correct proof is in the attachment

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