Question

simplify the expression
csc(-x)
1+tan^2(x)

Answer 1

Hint: for the first one. sin(x) is an odd function which means sin(-x)=-sin(x)

Answer 2

Hint: for the second one. Think about your Pythagorean identities.

Answer 3

So is it sin(x)tan(x)?

Answer 4

I'm not sure how you got csc(-x) is sin(x)tan(x)

Answer 5

recall \[\csc(x)=\frac{1}{\sin(x)}\]

Answer 6

I have no idea how to do this. lol

Answer 7

\[\csc(-x)=\frac{1}{\sin(-x)}=\frac{1}{?}\]
can you use anything I have mentioned here to rewrite sin(-x)?

Answer 8

The equation is over each other like csc(-x) over 1+tan^2(x) and the choices given are tan(x)
-cos(x)tan(x)
-cos(x)cot(x)
sin(X)tan(x)

Answer 9

ok so you meant to write a fraction

Answer 10

yes!

Answer 11

anyways the expression I was asking about can you rewrite it

Answer 12

i dare you to use the hint I gave above

Answer 13

1/sin(-x) =?

Answer 14

1 over sin(x) ?

Answer 15

sin(-x) doesn't equal sin(x)

Answer 16

sin(-x) equals -sin(x)
since sin(x) is an odd function

Answer 17

this was the first hint I mentioned

Answer 18

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

Answer 19

now can you use your Pythagorean identities to rewrite the bottom of your fraction as one term
that is I'm asking you to rewrite
1+tan^2(x)

Answer 20

is that sec^2(X)?

Answer 21

yes

Answer 22

so you have this so far:
\[\frac{\csc(-x)}{1+\tan^2(x)} \\ =\csc(-x) \cdot \frac{1}{1+\tan^2(x)} \\ =\frac{1}{\sin(-x)} \cdot \frac{1}{1 +\tan^2(x)} \\ = \frac{1}{-\sin(x)} \cdot \frac{1}{\sec^2(x)}\]

Answer 23

now recall that sec(x) and cos(x) are reciprocal functions of one another

Answer 24

that is \[\frac{1}{\sec(x)}=\cos(x)\]

Answer 25

try that to use that here

Answer 26

So -cos(x)tan(x)?

Answer 27

lets take little steps

Answer 28

1/sec^2(x)=?

Answer 29

cos^2(x)?

Answer 30

yes
so we have
\[\frac{1}{-\sin(x)} \cdot \cos^2(x) \\ \frac{1}{-\sin(x) } \cos(x) \cos(x) \]
now rearrange some things so you can see this is equivalent to one of your choices

Answer 31

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