Verified answer

(1 - cos) / (1 + cos) = (1-cos)^2 / (1+cos)(1-cos) = (1 - cos)^2/(1 - cos^2) = (1 - cos)^2/ sin^2 = [(1-cos)/sin]^2 = [ 1/sin - cos/sin ] ^2 =

[ csc - cot ] ^2

Multiply by the conjugate

(1-cosθ)(1-cosθ)

(1+cosθ)(1-cosθ)

(1-cosθ)^2

1-(cosθ)^2

(sin^2 + cos^2 = 1)

(1-cosθ)^2

1-[1-(sinθ)^2]

(1-cosθ)^2

(sinθ)^2

[(1-cosθ)/(sinθ)]^2

[1/sinθ - cosθ/sinθ]^2

[cscθ - cotθ]^2

Hi,

(1 - cos θ) / (1 + cos θ)

Mulitply the top and bottom by (1 - cos θ)

= (1 - cos θ)(1 - cos θ) / (1 + cos θ)(1 - cosθ)

= (1 - 2 cos θ + cos² θ) / (1 - cos² θ)

= (1 - 2 cos θ + cos² θ) / (sin² θ)

= (1/sin² θ) - 2(cos θ/sin θ)(1/sin θ) + (cos²θ/sin² θ)

= csc² θ - cot θ csc θ - cot θ csc θ + cot² θ

= (csc θ - cot θ)²

Hope this helps!

Well obviously the other people have already answered the question, but just a hint: use your Pythagorean Identites.

multiply top and bottom by [ 1 - cos Θ ] and use your knowledge of trig

cosx=a million/secx LHS=((a million/SECX)+a million)/((a million/SECX)-a million) now take LCM = ((a million+secx)/secx)/((a million-secx)/secx) now the denominator term secx gets cancelled you're left with =(a million+secx)/(a million-secx) =RHS hence the information

(1-cos@)/(1+cos@) =

(1-cos@)^2 / (1+cos@)(1-cos@) =

(1-cos@)^2 / 1-cos^2@ =

(1-cos@)^2 / sin^2@ =

(1-cos@/sin@)^2 =

(1/sin@ -cos@/sin@)^2 =

(csc@ -cot@)^2

(1-cos(θ)) / (1+cos(θ)) = (1-cos(θ))*(1+cos(θ)) / (1+cos(θ))^2 =

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

[sin(θ) / (1+cos(θ))]^2 =

[sin(θ)*(1-cos(θ)) / (1+cos(θ))*(1-cos(θ))]^2 =

[sin(θ)*(1-cos(θ)) / (1-cos(θ))^2 ]^2 =

[sin(θ)*(1-cos(θ)) / sin^2(θ) ]^2 =

[(1-cos(θ)) / sin(θ)]^2 =

[1/sin(θ) - cos(θ)/sin(θ)]^2 =

[csc(θ) - cot(θ)]^2