Question

Hi. I am reposting this. Can you verify. Thanks.

(cosθ) / (sinθ – 1) + (cosθ)/ (1 + sinθ) = -2tanθ



(1 / cos^2θ) – (2) (1/cos^2θ) (sinθ / cosθ) + (sin^2θ / cos^2θ) = (1 - sinθ) / (1 + sinθ)

Answer 1

ok

I usually start with the more complicated looking side and try to change it to the more simplified

LS = cosØ/(sinØ - 1) + cosØ(sinØ + 1) , note it didn't matter to change the order of that last denominator

common denominator of (sinØ-1) and (sinØ+1) is (sinØ-1)(sinØ+1)
ahhh, difference of squares!
result is sin^2 Ø -1 or -cos^2 Ø

= ( cosØ(sinØ - 1) + cosØ(sinØ + 1) )/((sinØ-1)(sinØ+1))
= (sinØcosØ - cosØ + sinØcosØ + cosØ)/-cos^2 Ø
= 2sinØcosØ/-cos^2 Ø
= -2sinØ/cosØ
= -2tanØ
= RS

the 2nd is a variation of one I did yesterday, let me try to find it

Answer 2

from last night at 10:06 pm

RS = (1-sinØ)/(1+ sinØ)
= (1-sinØ)/(1+ sinØ) * (1-sinØ)/(1- sinØ)
= (1 - 2sinØ + sin^2 Ø)/(1 - sin^2 Ø)
= (1 - 2sinØ + sin^2 Ø)/cos^2 Ø
= 1/cos^2 Ø - 2sinØ/cos^2 Ø + sin^2 Ø/cos^2Ø
= sec^2 Ø - 2(sinØ/cosØ)*(1/cosØ) + tan^2 Ø
= sec^2 Ø - 2tanØ secØ + tan^2 Ø
= 1/cos^2 Ø -2(sinØ/cosØ)(1/cosØ) + sin^2 Ø/cos^2 Ø
= your LS

Steve also gave you an alternate solution in that same post