Question

Simple, but hard math question:

Answer 1

Click to enlarge!

Answer 2

i will help you

Answer 3

well thank you

Answer 4

your welcome :)

Answer 5

divide every term by \(\sin^2\theta\)

Answer 6

and then tell me what you get

Answer 7

ok let me do that

Answer 8

sure, take your time

Answer 9

ok i got 1=1

Answer 10

???

Answer 11

if i did that correctly

Answer 12

ok let me try again

Answer 13

if you got 1=1 you did it correctly, but you went too far

Answer 14

I will direct you

Answer 15

oh

Answer 16

Your identity is: \(\large\color{slate}{\displaystyle \sin^2\theta+\cos^2\theta=1}\)
and you are given some toher options, and you have to choose one option equivalent to that.

I advised you to divide everything by \(\large\color{slate}{\displaystyle \sin^2\theta}\), and this is what I mean:

\(\large\color{red}{\displaystyle \frac{\color{black}{\sin^2\theta}}{\sin^2\theta}\color{black}{+}\frac{\color{black}{\cos^2\theta}}{\sin^2\theta}\color{black}{=}\frac{\color{black}{1}}{\sin^2\theta}}\)

Answer 17

and then simplify this expression, and tell me what is the correct option.

Answer 18

alright I got \[\csc ^{2} 0\]

Answer 19

\(\large\color{slate}{\displaystyle \large\color{red}{\displaystyle \frac{\color{black}{\sin^2\theta}}{\sin^2\theta}\color{black}{+}\frac{\color{black}{\cos^2\theta}}{\sin^2\theta}\color{black}{=}\frac{\color{black}{1}}{\sin^2\theta}}}\)

\(\large\color{slate}{\displaystyle ~~~~\Downarrow ~~+~~~\Downarrow~~~=~~\Downarrow}\)

\(\large\color{slate}{\displaystyle ~~~~~1~~+~~\cot^2\theta=\csc^2\theta}\)

Answer 20

i see. from ^ i choose B. i didnt realize that you meant solve the whole thing though

Answer 21

my bad

Answer 22

it's ol

Answer 23

ok