Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

Rewrite the expression in terms of the given function or functions. (sec x + csc x) (sin x + cos x) - 2 - cot x; tan x

OpenStudy (anonymous):

tanx is the given function ?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

thats how its written

OpenStudy (anonymous):

lets first look at : (secx + cscx)(sinx+cosx)

OpenStudy (anonymous):

and you should know that secx = 1/cosx and cscx = 1/sinx

OpenStudy (anonymous):

so can you simplify it ?

OpenStudy (anonymous):

what is cscx its not on my calc?

OpenStudy (anonymous):

i just wrote it :) cscx = 1/sinx

OpenStudy (anonymous):

wow sry loll

OpenStudy (anonymous):

so can you do it now ?

OpenStudy (anonymous):

1 sec

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i get sec(x) for that prob up there

OpenStudy (anonymous):

and im wrong bc thats not a answer to choose from

OpenStudy (anonymous):

lets do it together since i got something else. (secx + cscx)(sinx+cosx) = secxsinx + secxcosx + cscxsinx +cscxcosx = tanx + 1 + 1 + cotx

OpenStudy (anonymous):

so (secx + cscx)(sinx+cosx) = 2 + cotx + tanx now lets add it to the rest : (secx + cscx)(sinx+cosx) - 2 - cotx = 2 + cotx + tanx -2 -2 - cotx = tanx

OpenStudy (anonymous):

ill be back in some minutes i hope it is enough though

OpenStudy (anonymous):

i must not be puttin it i mcalc right

OpenStudy (anonymous):

2tan x??

OpenStudy (anonymous):

@Coolsector

OpenStudy (anonymous):

@britbrat4290 hey im sorry im back now

OpenStudy (anonymous):

as i wrote i got tanx

OpenStudy (anonymous):

shoot, ty

OpenStudy (anonymous):

yw .. say hey to the dolphin

OpenStudy (anonymous):

:)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!