Question

Help with checking work for using fundamental trig identities

Answer 1

If csc(θ) = -6, and θ is in the fourth quadrant, find cot θ.

Answer 2

-cot^2=csc^2+1

\[-\cot=\sqrt{(-6)^2+1} \]

\[-\cot=\sqrt{37}\]

\[\cot=-\sqrt{37}\]

Answer 3

\[1+\cot^2(x)=\csc^2(x)\]
did you use this identity?

Answer 4

Yes I just realized i didn't put -csc^2 so it should be -(6)^2

Answer 5

or should it just be positive actually

Answer 6

\[\cot^2(x)=\csc^2(x)-1 \text{ by subtracting one on both sides }\]

Answer 7

\[\cot(x)=\pm \sqrt{\csc^2(x)-1} \\ \text{ since we know we are \in the 4th quadrant } \\ \text{ we know } \cot(x) \text{ is negative } \\ \text{ so we actually have \in this case } \\ \cot(x)=- \sqrt{\csc^2(x)-1}\]

Answer 8

Man I really screwed up with the adding it over. I over complicated it again

Answer 9

ugh so it would be \[-\sqrt35\]

Answer 10

sounds great

Answer 11

At least its clicking just gotta fix some things

Answer 12

Hey freckles, for my next one i am dealing with sec and tan in the second quadrant. They are both negative so if we were dealing with the same question just switched out the trig functions would it be the same answer\[\tan(x)=-\sqrt{-\sec^2-1}?\]

Answer 13

because -sec would just change -6^2 to just 6^2

Answer 14

sounds like you have the following
we are in the second quadrant
-> this implies cosine is negative and sine is positive
->tan=sin/cos=+/-=-
So tan is negative
--
Not sure why you put a negative sign in front of the sec^2(x) part we never had to question if we need a neg or pos there

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\[\tan^2(x)+1=\sec^2(x)\]
\[\tan^2(x)=\sec^2(x)-1 \text{ by subtracting one on both sides }\]

\[\text{ now take square root of both sides }\]
\[\tan(x)=\pm \sqrt{\sec^2(x)-1}\]
We already said tan has to be negative because we are in the second quadrant so we know to drop the + and keep the -
so we have
\[\tan(x)=- \sqrt{\sec^2(x)-1}\]

Answer 15

recall
\[|x|=\sqrt{x^2}=x \text{ if } x>0 \\ \text{ or } -x \text{ if } x<0 \]

Answer 16

I put a -sec because it is negative in the 2nd quadrant though

Answer 17

is it wrong to do?

Answer 18

the only thing that has a question in front of it is sqrt(tan^2(x))

Answer 19

so it doesn't matter if if the other is negative or not?

Answer 20

\[\tan^2(x)=\sec^2(x)-1 \\ \\ \text{ take square root of both sides } \\ \sqrt{\tan^2(x)}=\sqrt{sec^2(x)-1}\\ \]
\[\text{ this actually means } |\tan(x)|=\sqrt{\sec^2(x)-1} \\ \text{ which means } \tan(x)= \sqrt{\sec^2(x)-1} \text{ or } -\sqrt{\sec^2(x)-1} \\ \text{ depending on what quadrant we are in }\]

Answer 21

this definitely does not give other options such as the one you presented:
\[\tan(x) \neq \pm \sqrt{\pm \sec^2(x)-1}\]

Answer 22

do you see where the or is occurring and why?

Answer 23

\[\sqrt{(f(x))^2}=f(x) \text{ if } f(x)>0 \\ \sqrt{(f(x))^2}=-f(x) \text{ if } f(x)<0\]
\[\tan^2(x)=\sec^2(x)-1 \\ \text{ here we take square root on both sides } \\ \text{ in attempt to solve for } \tan(x) \\ \sqrt{\tan^2(x)}=\sqrt{\sec^2(x)-1} \\ \text{ now remember } \sqrt{\tan^2(x)}=\pm \tan(x) \\ \pm \tan(x)=\sqrt{\sec^2(x)-1} \\ \tan(x)=\pm \sqrt{\sec^2(x)-1} \]
See there was no question ever that we should have also a plus or minus in front of the sec^2 term

Answer 24

now if we were solving for sec^2(x) given tan^2(x)
then we would have this instead:
\[1+\tan^2(x)=\sec^2(x) \\ \sqrt{1+\tan^2(x)}=\sqrt{\sec^2(x)} \\ \text{ but again} \sqrt{\sec^2(x)}=\pm \sec(x) \text{ depending on where we are exactly } \\ \sqrt{1+\tan^2(x)}=\sqrt{\sec^2(x)} \rightarrow \sqrt{1+\tan^2(x)}=\pm \sec(x) \\ \text{ so we have } \sec(x)=\pm \sqrt{1+\tan^2(x)}\]