Question

Prove that sinA-cosA+1/sinA+cosA-1=1/sec A-tan A?​

Answer 1

$\boxed {\boxed{ { \green{ \bold{ \underline{Verified \: Answer \: }}}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\bigstar{\underline{\underline{{\sf\ \red{ ANSWER:-}}}}}$

• REFER TO ATTACHMENT༒

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Answer 2

How do I prove sinA-cosA+1/sinA+cosA-1=1/secA-tanA?

Hi,

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

[math]LHS = (tanA - 1 + secA) / (tanA + 1 - secA)[/math]

Now

[math]sec^2 A = 1 + tan^2 A [/math]

[math]sec^2 A - tan^2 A = 1[/math]

Using above relation at denominator of LHS

[math]LHS = (tanA - 1 + secA) / (tanA - secA + sec^2 A - tan^2 A)[/math]

[math]LHS = (tanA - 1 + secA) / ((secA - tanA) (-1 + secA + tanA))[/math]

[math] LHS = 1 / (secA-tanA)[/math]

[math]LHS = RHS[/math]

Hence Proved.

I think above proof will clear your doubt,

All the best.