Question

A?

Answer 1

this calculation is taking me a while

Answer 2

I'm sorry i dont want you to have to do long calculations for me

Answer 3

>.<

Answer 4

so what I have so far is that you would multiply the num and denom csc(4x) - cot(4x), that'll convert the denominator into csc^2(4x) - cot^2(4x) which is 1 I believe

Answer 5

oh okay got what you are doing

Answer 6

I don't really like doing this but I can't think of a way to solve this w/o plugging in x values

Answer 7

Well, you could express them all in cos & sin terms and then simplify from there.

Answer 8

Like this:

\(\sf\dfrac{\frac{1}{sin(4x)}-\frac{cos(4x)}{sin(4x)}}{\frac{1}{sin(4x)}+\frac{cos(4x)}{sin(4x)}}\)

Answer 9

Combine the fractions in both the numerator and denominator:

\(\sf\dfrac{\frac{1-cos(4x)}{sin(4x)}}{\frac{1+cos(4x)}{sin(4x)}}\)

Answer 10

Use the division rule:

\(\sf\dfrac{(1-cos(4x))sin(4x)}{sin(4x)(1+cos(4x))}\)

Answer 11

Cancel out sin(4x):

\(\sf\dfrac{(1-cos(4x))\cancel{sin(4x)}}{\cancel{sin(4x)}(1+cos(4x))}\)

\(\sf\dfrac{1-cos(4x)}{1+cos(4x)}\)

Answer 12

and get b ?

Answer 13

Not quite, try again.

Answer 14

D

Answer 15

because of the 4x

Answer 16

We could do:

\(\sf\dfrac{1-cos(2\cdot 2x)}{1+cos(2\cdot 2x)}\)

Answer 17

C :/

Answer 18

Ohhh, then we could multiply 1/2 to the numerator and denominator, then use the identities:

\(\sf sin^2(x)=0.5(1-cos(2x))\)
\(\sf cos^2(x)=0.5(1+cos(2x))\)

Answer 19

Which leaves us with:

\(\sf\dfrac{sin^2(2x)}{cos^2(2x)}\)

\(\sf tan^2(2x)\)

\(\Huge\checkmark\)

Answer 20

thank you so much I really do appreciate it

Answer 21

No problem.