Question

Please can someone help? (cos(θ))/(1-sin(θ))-(1)/(cos(θ))=tanθ.

Answer 1

ok. lets make it easier -

\[\frac{ \cos x }{ 1-\sin x } - \frac{ 1 }{ \cos x }\]

Answer 2

where theta = x.

Answer 3

Multiplying cos x/1-sin x with cos x, and 1/cos x with cos x/1-sin x u get -

\[\frac{ \cos^2x-1 }{ (1-\sin x)\cos x }\]

Answer 4

okay. I follow.

Answer 5

w8 there's a bit error. the correct one is -

\[\frac{ \cos^2x - (1-\sin x)}{ (1-\sin x) \cos x }\]

Answer 6

this is the correct one.

Answer 7

so next step would be -

\[\frac{ \cos^2x - 1 + \sin x}{ (1-\sin x)\cos x }\]

Answer 8

now we know that 1 = cos^2x + sin^2x.

u can use that. hence we get -

\[\frac{ \cos^2x - (\cos^2x + \sin^2x) + \sin x }{ (1-\sin x)\cos x }\]

Answer 9

oh okay, sorry, I see what I was doing wrong now. You know how you made that little mistake, I did that too.

Answer 10

Haha right. :P

so now on taking out from brackets, we have -

\[\frac{ \cos^2x - \cos^2x -\sin^2x + \sin x }{ (1-\sin x)\cos x }\]

Answer 11

and cos^2x - cos^2x = 0, so you're left with sin x - sin^2x.

hence -

\[\frac{ \sin x - \sin^2x }{(1-\sin x)\cos x }\]

Answer 12

u can take sin x to be common and write it as - sin x(1-sin x)

\[\frac{ \sin x(1-\sin x) }{ (1-\sin x)\cos x }\]

Answer 13

You can cancel the common (1-sin x), following which we have -

\[\frac{ \sin x }{ \cos x } => Tan x. \]

Answer 14

Wow thank you so much. I don't think I could ever figured it out. I struggle so much with trig identity. :/ Thank you again.

Answer 15

haha don't worry. Initially even i used to hate them and all but later this became a very entertaining thing for me :P

Answer 16

Yeah, I can get that. They are fun to do, but I always get stuck, especially with the ones that are fractions and you have to + or -. :( I guess practice makes perfect. :)