Question

anyone care to tackle this?

Answer 1

*Breathes in* Nope D:

Answer 2

yeah i should hope not
it makes no sense at all
it is not an "equation" there is no equal sign
also the expression is the same as \(1+\sin(x)\) so who knows what they are after

Answer 3

D; looks like trig on crack O_o

Answer 4

Do you know how to solve?, if we can :-/

Answer 5

It's solvable... *heavy breath* it's very long way to go...

Answer 6

it needs basic trigometric identity and double angle as well to solve it.

Answer 7

I am in algebra 1, if this is anything beyond that i won't know it, but I won't shy away from learning how to do this kinda numbers and letters and angles and things

Answer 8

All you need to know to solve the question, it's all here. I am sorry but consider the time I have, I won't be able to go details of this with you.
http://www.purplemath.com/modules/idents.htm

Answer 9

Well, I never thought of this before but why not,

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

So...

\[\frac{\cos^2 \theta}{1-\sin \theta} = 1+\sin \theta \]

Uhhhh nothing makes any sense lol

Answer 10

Like why is the third option in a different font lol

Answer 11

\[(1+cosec)/cosec \\=(1/cosec)+1\\=1+\sin\]

Answer 12

can u teach me how to "tackle" this @Kainui I'm not great at tackling things, I'm kind of skinny

Answer 13

Come to think of it, this looks like two questions badly copy pasted together. If you look carefully you'll notice the text of the first part of the question up to the word "equation" is the same font as the 3rd answer and then the second half of the question along with the other 3 options are a separate font too.

Answer 14

I think @satellite73 just made this up, it's not even a real question to begin with lol.

Answer 15

oh

Answer 16

That, or someone is trolling @satellite73 lolol

Answer 17

i'm not sure i understand the question..
isn't it just simply the expression till you get one of the options?

Answer 18

or maybe satellite is smarter than us or he made up a rule or something

Answer 19

simplify*

Answer 20

just use the fill tool to see that there's a slightly different shade of white in the background where the other question was pasted in lolol

http://i.imgur.com/qKEf9zx.png

Answer 21

common core I assume

Answer 22

@Kainui I have seen this before in online curriculum where there are different fonts in the answers. Don't know why this is though, and I've seen it confuse students.

Answer 23

\(\color{brown}{\text Q}\)\[\color{brown}{\frac{\cos^2}{1-\sin\theta}=\frac{1-\sin^2\theta}{1-\sin\theta}=1+\sin\theta}\]
\(a)\)\[\frac{1-\sin\theta}{\cos\theta}=(1-\sin\theta )\sec\theta=\sec\theta-\tan\theta\]\(b)\)\[\frac{1+\sec\theta}{\sec\theta}=\cos\theta+\tan\theta\]\(c)\)\[\sin\theta\]\(\color{red}{\text d)}\)\[\color{red}{\frac{1+\csc\theta}{\csc\theta}=\sin\theta+1}\]

Answer 24

\[\frac{\cos^2 \theta}{1 - \sin \theta} \ne 1 + \sin \theta \]

Answer 25

it's the best option @ParthKohli

Answer 26

Why is this not true? \[\frac{\cos^2 \theta}{1 - \sin \theta} \ne 1 + \sin \theta \]

Part of me wants to say it's cause at \(\theta = n \pi+\frac{\pi}{2}\) the expression \(1-\sin \theta=0\) however since \(\cos \theta = 0\) at those exact same values we can use L'Hopital's rule to look at the indeterminate form and see that:

\[\lim_{\theta \to \pi/2} \frac{\cos^2 \theta}{1-\sin \theta} =\lim_{\theta \to \pi/2} \frac{-2\cos \theta \sin \theta}{-\cos \theta} =2 = 1 + \sin(\pi /2)\]

Answer 27

\(\frac{\cos^2\theta}{1-\sin(\theta)} - \boldsymbol{\sin \theta = (1 + \sin\theta) - \sin\theta} = 1\) ??? Does this mean I "completed the expression to obtain an identity". lol

Answer 28

the equation*

Answer 29

@Kainui I'd guess it's not true because it's not defined for all values of theta on the left. To hell with L'hopital and his rules.

Also http://openstudy.com/study#/updates/5760e4a1e4b0008ef3e077e6