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Calculus1 15 Online
OpenStudy (please.help.me):

Verify sec x(csc x − 2 sin x) = cot x − tan x ?

OpenStudy (thatonegirl_):

What is sec= to?

OpenStudy (please.help.me):

1/cos?

OpenStudy (thatonegirl_):

Yup! Now try distributing that into the parenthesis

OpenStudy (please.help.me):

sec x(csc x − 2 sin x) = 1/cosx (1/sinx - 2sin^2 x/?)

OpenStudy (please.help.me):

OpenStudy (johnweldon1993):

That's a terrible way to do it -_- But it works...Just look at that first unknown box...what do you think goes there? To make it match up with the left side

OpenStudy (johnweldon1993):

*Big hint there in the next step, you can see inside the parenthesis have been put over a common denominator*

OpenStudy (please.help.me):

sin(x) for the next two blanks?! Thank you!

OpenStudy (johnweldon1993):

Correct!

OpenStudy (please.help.me):

I'm confused how to get the third blank though

OpenStudy (johnweldon1993):

And the third box...they just multiplied in the 1/cos(x) into the parenthesis So we know 1 * the numerator is just the numerator But now that we know sin(x) is on the bottom...and that is being multiplied by cos(x)...what is our new denominator?

OpenStudy (johnweldon1993):

*Don't overthink it! :)

OpenStudy (please.help.me):

Is this an identity? Sorry, I am really bad at these problems because we just started with this in our class

OpenStudy (johnweldon1993):

Nope, no identity Haha reason why i said don't over think it :) What is cos(x) * sin(x)? Literally just cos(x)sin(x) That is your new denominator!

OpenStudy (sshayer):

\[\sec x \left( \csc x-2\sin x \right)=\sec x \left( \frac{ 1 }{ \sin x }-2\sin x \right)\] \[=\sec x \left( \frac{ 1-2\sin ^2x }{ \sin x } \right)=\frac{ \cos ^2x+\sin ^2x-2\sin ^2x }{ \cos x \sin x }\] \[=\frac{ \cos ^2x-\sin ^2x }{ \cos x \sin x }=?\]

OpenStudy (johnweldon1993):

Lol yes as @sshayer has put above in a neater fashion :)

OpenStudy (please.help.me):

Oh wow! Lol okay thank you! So the 4th blank is the same?

OpenStudy (johnweldon1993):

So those 3,4 and 5 boxes should be easy for you to fill in

OpenStudy (johnweldon1993):

Indeed boxes 3 and 4 are the same

OpenStudy (johnweldon1993):

Box 5 is quite simple as well...

OpenStudy (please.help.me):

Got it! Thank you so much! I am so glad I found out about this website! Is it free all the time or is there like a trial first?

OpenStudy (johnweldon1993):

Free all the time! There was a system in place before where you could pay to get "Qualified helper assistance" but that has sense disappeared *For the moment* But yeah, it's a great resource for help!

OpenStudy (johnweldon1993):

^Ignore my bad grammar there XD *sense --> Since*

OpenStudy (please.help.me):

Awesome! Thanks again!!)

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