Question

I need help proving trig equations! Help please?!

cot x sec^4x = cot x + 2 tan x + tan^3x

Answer 1

Okay, any attempts made?

Answer 2

I don't really know what I'm suppose to do, well I know but It's like confusing. I know I have to start on the left side but I don't understand what to do from there.

Answer 3

I think your supposed to take the side with more terms, and then try to get to the other side from there.

Answer 4

Oh, well the lesson made a big deal about always working from the left but I guess that makes sense. I have the Pythagorean identities. Would I be using that?

Answer 5

It's been about 2 almost 3 years since I took trig, I honestly don't remember.

Answer 6

take cotx common from R.H.S

Answer 7

\[\cot x \left( 1+\tan ^{2}x+\tan ^{4}x \right)\]

Answer 8

Would I be factoring? Im lost, sorry.

Answer 9

correction write 2tan^2 x in place of tan^2x

Answer 10

\[=\cot x \left( 1+\tan ^{2}x \right)^{2}\]

Answer 11

Have you followed?

Answer 12

I get the first part but from there what exactly is there to do?

Answer 13

\[1+\tan ^{2}x=\sec ^{2}x\]

Answer 14

you know the identity
\[\sin ^{2}x+\cos ^{2}x=1\]
\[divide~ each~ term~ by~ \cos ^{2}x\]

Answer 15

\[\cot x \left( \sec ^{2}x \right)^{2}=L.H.S\]

Answer 16

If not clear let me know.

Answer 17

This is all I've had written down

1. 1/tan x + 2 tan x + tan^3 x

1 + 2 tan x + 2 tan + tan^3 x / tan x

1 + tan^2 x + tan^2 x + tan^3 x / tan x

Answer 18

ignore the (1.)

Answer 19

So I'm dividing everything on the right side of the equation by cos^2x?

Answer 20

R.H.S\[=\cot x+2\tan x+\tan ^{3}x=\cot x+2\tan x \frac{ \cot x }{\cot x }+\tan ^{3}x \frac{ \cot x }{\cot x }\]
\[=\cot x \left( 1+2\frac{ \tan x }{ \cot x }+\frac{ \tan ^{3}x }{\cot x } \right)\]
\[=\cot x \left( 1+2\tan x \tan x+\tan ^{3}x \tan x \right)\]
\[=\cot x \left( 1+2\tan ^{2}x+\tan ^{4}x \right)\]
now you can complete.

Answer 21

\[=\cot x \left( 1+\tan ^{2}x \right)^{2}=\cot x \left( \sec ^{2} x\right)^{2}=\cot x \sec ^{4}x\]

Answer 22

cotx(1+tan2x+tan4x)

=cotx(1+tan2x)2

=cotx+2tanx+tan3x=cotx+2tanxcotxcotx+tan3xcotxcotx

=cotx(1+2tanxcotx+tan3xcotx)

=cotx(1+2tanxtanx+tan3xtanx)

=cotx(1+2tan2x+tan4x)

=cotx(1+tan2x)2=cotx(sec2x)2=cotxsec4x

^like that? Is the full operation?

Answer 23

you can also solve from L.H.S

Answer 24

Could you help me with another problem?

Answer 25

\[\cot x \sec ^{4}x=\cot x \left( \sec ^{2}x \right)^{2}=\cot x \left( 1+\tan ^{2}x \right)^{2}\]
\[=\cot x \left( 1+2\tan ^{2}x+\tan ^{4}x \right)\]
\[=\cot x \left( 1+2\tan x \tan x+\tan ^{3}x \tan x \right)\]
\[=\cot x \left( 1+\frac{ 2\tan x }{\cot x } +\frac{ \tan ^{3}x }{\cot x }\right)\]
\[=\cot x+2\tan x+\tan ^{3}x \]

Answer 26

i will try.

Answer 27

2. (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 28

\[\sin x \left( \tan x \cos x-\cot x \cos x \right)=\sin x \left( \frac{ \sin x }{\cos x }\cos x-\frac{ \cos x }{ \sin x }\cos x \right)\]
\[=\sin ^{2}x-\cos ^{2}x=1-\cos ^{2}x-\cos ^{2}x=1-2\cos ^{2}x\]

Answer 29

Thank you! Could you help me with the one question Im stuck on now? I'll be able give you a medal on there. http://openstudy.com/study#/updates/52d59e57e4b0274b88c513fb