Question

*Will fan and medal*
The expression BLANK is NOT equivalent to (1 − sin2(x)) tan(-x).

A) (1-cos^2(x)) cot(-x)

B) (cos^2(x)-1) cot(x)

C) (sin^2(x)-1) tan(x)

D) (cos^2(x)-1) cot(-x)

Answer 1

Wait, is the QUESTION supposed to be (1 - sin^2(x))(tan(-x))

Answer 2

I just copied what the question was, but maybe it can be written as that also can it?

Answer 3

Yeah what you wrote is fine.

Answer 4

So lets see

\[\large (1 - sin^2(x))(tan(-x))\]

Frst, we know tan is an odd function so \(\large tan(-x) = -tan(x)\)
\[\large (1 - sin^2(x))(-tan(x))\]

We know \(\large tan(x) = \frac{sin(x)}{cos(x)}\)
so

\[\large (1 - sin^2(x))(-\frac{sin(x)}{cos(x)})\]
And finally we know \(\large 1 - sin^2(x) = cos^2(x)\) so

\[\large (cos^2(x))(-\frac{sin(x)}{cos(x)})\]
Leaving

\[\large -cos(x)sin(x)\]

Now we just need to find if any of your options simplify down to this

Answer 5

ok I see

Answer 6

Want me walk through it with ya or do you think you got it? :)

Answer 7

walk me through it please

Answer 8

Okay :) so it will all be the same exact steps

A) \(\large (1-cos^2(x))(cot(-x)\)

Now we know cot is also an odd function...so again \(\large cot(-x) = -cot(x)\) and we also still know that \(\large 1 - cos^2(x) = sin^2(x)\)

So lets rewrite what we have so far

\[\large (sin^2(x))(-cot(x))\]

Now we know \(\large cot = \frac{cos(x)}{sin(x)}\)
so

\[\large (sin^2(x))(-\frac{cos(x)}{sin(x)})\]
Leaving us with
\(\large -cos(x)sin(x)\)

So we know that's not the answer

Answer 9

alright I see

Answer 10

So go ahead and try out B and tell me what you get :)

Answer 11

I'm retarded help

Answer 12

I don't know how to type it in my calculator because it wont let me square cos or any of those

Answer 13

Nope you dont even need to do any of that...

\[\large 1 - cos^2(x) = sin^2(x)\]
and
\[\large tan(x) = \frac{sin(x)}{cos(x)}\]

are just trig identities, VERY important to memorize

Answer 14

ok I see

Answer 15

Yeah absolutely no calculator required with these :)