Question

MATH HELP !
(Picture below)

Answer 1

@satellite73

Answer 2

@.Sam.

Answer 3

you cannot divide by zero, and sine is zero at
\[0,\pi,2\pi,3\pi, ...\]

Answer 4

go with the last option

Answer 5

So its d ?

Answer 6

yea

Answer 7

Can you help me with more @satellite73

Answer 8

I only have 3 more questions after this that i need help with.

Answer 9

\[\csc(x)=\frac{1}{\sin(x)}\] so
\[\cos(x)\times \csc(x)=\cos(x)\times \frac{1}{\sin(x)}=\frac{\cos(x)}{\sin(x)}=\cot(x)\]

Answer 10

your answer is "b"

Answer 11

would it be b

Answer 12

Okay thank you sooo much. I have 3 more that im stuck on

Answer 13

ok i can look at them

Answer 14

ia it tangent identity ?

Answer 15

@satellite73

Answer 16

You there ?

Answer 17

yeah i am not sure what you are supposed to put for this one
they all can be used except maybe for the silly \(\csc^2(x)=\csc^2(x)\)

Answer 18

My answer choice

Answer 19

oooh i see
i thought those were your choices
i would say "tangent identity"

Answer 20

Lol are you sure ?

Answer 21

pretty sure, yes
since it is not used anywhere

Answer 22

1 more after this one lol Please & Thank you soo much

Answer 23

the third one is for sure

Answer 24

none of the rest are

Answer 25

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

Answer 26

for that one when you add you get
\[\frac{1+\sin(x)+1-\sin(x)}{1-\sin^2(x)}=\frac{2}{\cos^2(x)}=2\sec^2(x)\]

Answer 27

Thank you soooo much :)