At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

i can't scroll..

Try refreshing the page, or log out and log back in.

sinx cosx/tanx=sinx cosx cosx/sinx
=cosxcosx
=cos^2x
=1-sin^x

sorry...1-sin^2x

how does sinxcosx/tanx = sinx?

It doesn't. It equals cos^2x

just kidding, i understand how you wrote it now

I like how you stalked me then...lol

haha sorry. i got confused.

1 / sec(x) - 1 minus 1 / sec(x) + 1 = 2cot^2(x)

Can you type out the question using the equation editor?

\[1\div \sec(x-1) - 1 \div \sec(x+1) = 2\cot^2x\]

okay

in the denominator it's supposed to be sec(x) - 1 and sec(x) + 1

Yeah...I was thinking that.

yeahh.

Okay, it's easy now...

\[=2\cot^2 x\]

Then cot^2(x) is just from the definition, \[\cot x = \frac{1}{\tan x}\]

Once you're done, let me know if it works for you or not.

okay, i understand that one

\[\cos(\pi/2 - x)\cos(-x) = \sin(2x) \div 2\]

how did you identify (1)?

is x. I'll draw a pic.

yeah, i get it now. i just had to go through it all

Okay

\[\tan(x \div 2) \cos^2(x) - \tan(x \div 2) = \sin(x) \div \sec(x) minus \sin (x)\]

sin(x) is subtracted from sin(x) / sec(x) in the second equation.

Okay, sorry, had to do something. Here we go...

ok, yep.

got it so far..

Forget the tan^2(x/2) thing I started with - transcribing notes to this site leads to screw-ups.

ok

find exact value of (-pi/12)

function?

sin(-pi/12)

sorry

square root of 3 over 4?

Do you want to try that and see how you go. I need a shower.

You can read off your sine, cosine and tangents from them.

You could also use the combination,\[\frac{\pi}{4}-\frac{\pi}{3}\]