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

cot x sec^4 (x) = cot x + 2 tan x + tan^3 (x) how do i make these equal and please show all the steps because i want to know how to do this

OpenStudy (larseighner):

What is sec^2 equal to?

OpenStudy (anonymous):

isnt it 1+tan^2x

OpenStudy (larseighner):

Right. So substitute that twice for sec^4 x.

OpenStudy (anonymous):

ok

OpenStudy (larseighner):

In other words you have (1 + tan^2s)^2 (and you still have to mutiply that by cot x; So expand (1+ tan^2)^2

OpenStudy (anonymous):

so it would be (cotx)(1+tan^2x)(1+tan^2x)

OpenStudy (larseighner):

Right, but multiply the 1+ tan^2x terms by each other. If you cannot figure that out, just expand (1+a)^2 and the substitute tan^2 for a when you are done.

OpenStudy (anonymous):

ok now what

OpenStudy (larseighner):

Then multiply by the cot x. And that should give you the answer.

OpenStudy (anonymous):

? im confused

OpenStudy (larseighner):

How so? Do you not know what (cot x)(tan x) is?

OpenStudy (anonymous):

well this is what i have now (cotx)(1+tan^2x)(1+tan^2x)=cotx+2tanx+tan^3x

OpenStudy (larseighner):

Isn't that what you were trying to prove?

OpenStudy (anonymous):

yes

OpenStudy (larseighner):

Well, you've done it.

OpenStudy (anonymous):

i dont see how

OpenStudy (larseighner):

\[ \large \begin{align} &\cot x \sec^4 (x) \cr &= (\cot x)(1+\tan^2x)(1+\tan^2x) \cr &= (\cot x)(1+2\tan^2x +\tan^4x) \ \cr &= \cot x+2\tan x +\tan^3x \end{align}\] That is what you did.

OpenStudy (larseighner):

You took sec^4 x to be sec^2 x sec^2x, You substituted (1 + tan^2x) for each occurrence of sec^2 x. Then you multiplied the terms (1 + tan^2). Then you multiplied by cot x. Since cot x is the reciprocal of tan x, cot x canceled one tan x out of each term with a tan x.

OpenStudy (anonymous):

ohhh ok i get it now

OpenStudy (anonymous):

thanks

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!