proving identities in trigonometry...?

Question

prove that (sin a - cos a - 1)/(sin a +cos a - 1) = - (cos a + 1)/sin a

PLEASE HELP, I'm so overloaded with homework, I cannot focus on this easy question... please help...

Answer 1

Notice that both numerator and denominator hav sin a - 1 common, then,

[(sin a - 1) - cos a] / [(sin a -1) + cos a]
Multiplying top and bottom by the conjuguate of the denominator we get,
[(sin a - 1) - cos a]^2 / [(sin a -1)^2 - cos^2 a]
I think you can work it there on

Answer 2

(sin a - cos a - 1)/(sin a +cos a - 1) = - (cos a + 1)/sin a

multiply both sides with with sin a:
=>(sin a)(sin a - cos a - 1)/(sin a +cos a - 1) = - (cos a + 1)

multiply both sides with with (sin a +cos a - 1):
=>(sin a)(sin a - cos a - 1) = - (cos a + 1)(sin a +cos a - 1)

=> (sin a)^2 - sin a cos a - sin a = - sin a cos a - (cos a)^2 + cos a - sin a - cos a + 1
=> (sin a)^2 - sin a cos a - sin a = - sin a cos a - (cos a)^2 - sin a + 1
=> (sin a)^2 = - (cos a)^2 + 1 (add sin a cos a + sin a to both sides of the equation)
=> (sin a)^2 + (cos a)^2 = 1
=> 1 = 1 (using the rule (sin a)^2 + (cos a)^2 = 1)

qed

Answer 3

= ((sin a - 1) - cos a) / ((sin a - 1) + cos a)

= ((sin a - 1) - cos a)* ((sin a - 1) - cos a) /
((sin a - 1) + cos a)* ((sin a - 1) - cos a)

Try mutiplying with the conjugate. It is difficult working it on the keyboard.