Question

sinA-cosA-1÷sinA+cosA-1=1÷secA-tanA​

Answer 1

$\bold{Given:-}$

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

$\bold{To prove:-}$

• LHS=RHS

$\bold{Solution:-}$

• $\frac{sinA-cosA+1}{sinA+cosA-1}$

• divide by cos A

• as we know
• $\frac{sinA}{cosA}$=tanA

• $\frac{1}{cosA}$=secA

• $\frac{tanA-1+sec A}{tanA+1-secA}$

• sec²A-tan²A=1

• $\frac{tanA-(sec²A-tan²A)+sec A}{tanA+1-secA}$

• as we know a²-b²=(a-b)(a+b)

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

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

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

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

• tanA+secA

• rationalise

• $\frac{tanA+secA}{secA-tanA}$×secA-tanA

• as we know a²-b²=(a-b)(a+b)

• $\frac{sec²A-tan²A}{secA-tanA}$

• sec²A-tan²A=1

• $\frac{1}{secA-tanA}$

hence ,proved

LHS=RHS

Answer 2

Answer:

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