Can you please prove the following identities?

Question

A) (1-sinΘ) / (cosΘ) = (cosΘ) / (1+sinΘ)

B) (1-2sin²Θ) / (cosΘ+sinΘ) = (cosΘ - sinΘ)

Answer 1

a) (1-sinΘ) / (cosΘ) = [(1-sinΘ)(1 + sinΘ)] / [((cosΘ)(1 + sinΘ)] = (1 - sin^2Θ)/ [((cosΘ)(1 + sinΘ)] = (cos^2Θ)/[((cosΘ)(1 + sinΘ)] = (cosΘ) / (1+sinΘ) (for Θ <> k pi + pi/2)


b)(1-2sin²Θ) / (cosΘ+sinΘ) = (cos^2Θ + sin^2Θ - 2sin²Θ)/(cosΘ+sinΘ) = (cos^2 - sin^2Θ)/(cosΘ+sinΘ) = (cosΘ + sinΘ)(cosΘ - sin Θ)/(cosΘ+sinΘ) = cosΘ - sinΘ (for cosΘ + sinΘ <> 0 => Θ <> k pi + 3pi/4)

Answer 2

a) (1-sinΘ) / (cosΘ) = (cosΘ) / (1+sinΘ)
working on the right hand side
(cosΘ) / (1+sinΘ) (1-sin Θ/1-sin Θ)
[cos Θ(1-sin Θ)]/(1-sin^2 Θ)
recall,
sin^2 Θ+cos^2 Θ =1
cos^2 Θ= 1-sin^2 Θ
then
cos Θ (1-sin Θ)/cos^2 Θ
(1-sin Θ)/cos Θ

Answer 3

A) cos^2(Θ)=(1-sinΘ)(1+sinΘ)

cos^2(Θ)=(1^2)-(sin^2(Θ))
cos^2(Θ)+sin^2(Θ)=1
which is true...

B) Just do the same thing with the other one

Just keep in mind that cos^2(Θ)+sin^2(Θ)=1...

Answer 4

Both are simple results of cos^2Θ + sin^2Θ = 1.

(A) cos^2Θ = 1 - sin^2Θ = (1-sinΘ)(1+sinΘ)
Divide both sides by (1+sinΘ)cos(Θ) gives you your result.

(B) 1-2sin^2Θ = cos^2Θ + sin^2Θ - 2sin^2Θ
= cos^2Θ - sin^2Θ = (cosΘ+sinΘ)(cosΘ-sinΘ)

Divide both sides by cosΘ+sinΘ.

Answer 5

gosh what are you some kinda genius????? Who in the world knows how to do these problems???? Im struggle-ing with geomoetry!!!!!!!! Bythe way if you are really smart in math e-mail me cause i need some serious help in that area!!!!!! But god luck trying to figure out those problems!!!!!!

Answer 6

A) (1-sin)/cos
(1+sin)(1-sin) / [cos(1+sin)] <--- multiply top/bottom by 1+sin
(1-sin^2) / [cos(1+sin)] <--- FOIL the top
cos^2 / [cos(1+sin)] <--- sin^2 + cos^2 = 1 identity on the top
cos / (1+sin) <--- cancel a cos from top/bottom

B) (1-2sin^2) / (cos+sin)
[(sin^2+cos^2)-2sin^2] / (cos+sin) <--- sin^2 + cos^2 = 1 identity on the top
(cos^2-sin^2) / (cos+sin) <--- subtract on top
(cos+sin)(cos-sin) / (cos+sin) <--- FOIL top
cos-sin <--- cancel (cos+sin) from top/bottom

Answer 7

Yes I can.