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

Verify the identity cot(x-pi/2) = -tan x

OpenStudy (anonymous):

please help

OpenStudy (shamim):

cot(-(π/2-x)) right?

OpenStudy (anonymous):

no cot(x-pi/2)=-tanx

OpenStudy (shamim):

i took left hand side

OpenStudy (anonymous):

explain please

OpenStudy (anonymous):

how to get the answer..i don't understand this

OpenStudy (shamim):

cot(x-π/2)=cot(-(π/2-x)=-cot(π/2-x) right?

OpenStudy (anonymous):

yeah what would be the answer

OpenStudy (shamim):

=-tanx=right hand side

OpenStudy (anonymous):

so it would be cot(-pi/2-x)

OpenStudy (anonymous):

sorry im really bad at math

OpenStudy (anonymous):

someone please help

OpenStudy (anonymous):

anyone!!!!

OpenStudy (shamim):

x-π/2=-(π/2-x) right?

OpenStudy (anonymous):

why does it equal it can you please explain

OpenStudy (shamim):

a-b=-(b-a) right?

OpenStudy (anonymous):

ok yes

OpenStudy (shamim):

did u really get it?

OpenStudy (anonymous):

yes because in each thing a will be pos and b neg right? so that's why they are equal

OpenStudy (anonymous):

plase help

OpenStudy (anonymous):

please

OpenStudy (shamim):

so why not x-π/2=-(π/2-x)

OpenStudy (anonymous):

awww I get that part

OpenStudy (shamim):

ok very good

OpenStudy (shamim):

do u know sin(π/2-30)=cos30

OpenStudy (anonymous):

yes

OpenStudy (shamim):

ok

OpenStudy (anonymous):

which identity would i use:)

OpenStudy (shamim):

cot(π/2-x)= tan x right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

that true that's an identity i didn't see that..but how would you add the neg

OpenStudy (shamim):

now sin(-x)=-sin x right?

OpenStudy (anonymous):

yes

OpenStudy (shamim):

tan(-x)=-tan x right

OpenStudy (anonymous):

yes

OpenStudy (shamim):

in similar way cot(-x)=-cot x right?

OpenStudy (anonymous):

yes

OpenStudy (shamim):

sowe can write cot(-(π/2-x))=-cot(π/2-x) right?

OpenStudy (anonymous):

yes you can. but since the problem is cot(x-pi/2)....how would I add the neg since I don't think it can be changed to -cot(-x+pi/2) right?

OpenStudy (shamim):

but we can write cot(x-π/2)=cot(-(π/2-x)) because x-π/2=-(π/2-x)

OpenStudy (shamim):

right?

OpenStudy (anonymous):

ohh okay I see.

OpenStudy (anonymous):

so would it mean cot(-(pi/2-X) = -cot(pi/2-x)..like you said before

OpenStudy (shamim):

ya u r correct

OpenStudy (anonymous):

so -cot(pi/2-x) = -tanx right?

OpenStudy (shamim):

ya

OpenStudy (anonymous):

ohh okay, Thank you so much for your help!!

OpenStudy (shamim):

welcome

OpenStudy (shamim):

anyway which grade u r

OpenStudy (anonymous):

12 grade

OpenStudy (anonymous):

thank you so much again you are truly a life saver

OpenStudy (shamim):

which country?

OpenStudy (shamim):

i wanna solve ur every math if i can collect ur math book

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!