Question

Prove that (sin θ − cos θ + 1) / (sin θ + cos θ − 1) = 1 / (sec θ − tan θ) using the identity sec² θ + tan² θ = 1.

Answer 1

Hi ,

Here I'm taking ' A ' instead of theta.

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

divide numerator and denominator

with cosA , we get

= ( tanA-1+secA)/(tanA+1-secA)

=( tanA+secA-1)/(tanA+1-secA)

=[(tanA+secA+(sec²A-tan²A)]/(tanA+1-secA)

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

= {(tanA+secA)[1 -(secA-tanA )]}/(tanA+1-secA)

= [(tanA+secA)(1-secA+tanA)](1-secA+tanA)

after cancellation ,

= tanA + secA

= [( secA + tanA )(secA - tanA )]/( secA-tanA)

= ( sec²A - tan²A )/( secA - tanA )

= 1/( secA - tanA )

= RHS

I hope this helps you.

: )