Question

Please help me!!
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

cot x sec4x = cot x + 2 tan x + tan3x

Answer 1

is it \(\cot(x)\sec^4(x)\)?

Answer 2

@satellite73 yes!!!!

Answer 3

ok then the left hand side can be rewritten as
\[\cot(x)\sec^4(x)=\csc(x)\sec^3(x)\] since \(\cot(x)=\frac{\cos(x)}{\sin(x)}\) as one of the cosines cancel
now lets see what we can do to the right hand side

Answer 4

@satellite73 okay so the right side will be {cos(x)/sin(x)}+2tanx+tan^3(x)??
btw does that sat sec to the second power or third power?

Answer 5

this is a raft of algebra

Answer 6

let me see if i can straighten in out clearly before i start writing

Answer 7

the idea will be to replace \[2\tan(x)+\tan^3(x)=\tan(x)(2+\tan^2(x))=\tan(x)(1+1+\tan^2(x))\]
\[=\tan(x)(1+\sec^2(x))\] and then add the fractions

Answer 8

@satellite73 could you work out the whole solution so i could get an idea of it?O.o

Answer 9

yes i will as soon as i get it
you are sure there is not a typo here right? because i am getting very close, but not getting it exactly

Answer 10

@satellite73 hmm what typo could have been made?cot x sec^4(x) = cot x + 2 tan x + tan^3(x) does this look better??:)

Answer 11

ok i am not getting it exactly
in fact i am not sure it is right
lets do some algebra

Answer 12

@satellite73 okay!!

Answer 13

it is definitely
\[\cot(x)\sec^4(x)=\cot(x)+2\tan(x)+\tan^3(x)\] right?

Answer 14

@satellite73 yupp

Answer 15

ok lets replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) to make the algebra simpler
then if we see \(a^2+b^2\) we can replace it by 1 since \(\cos^2(x)+\sin^2(x)=1\)

Answer 16

the left side is
\[\frac{a}{b}\times \frac{1}{a^4}=\frac{1}{a^3b}\]

Answer 17

@satellite73 yea

Answer 18

ok now before we start on the right hand side, lets note that
\[2\tan(x)+\tan^3(x)=\tan(x)\left(2+\tan^2(x)\right)=\tan(x)\left (1+1+\tan^2(x)\right)\]
\[=\tan(x)\left(1+\sec^2(x)\right)\] since \(1+\tan^2(x)=\sec^2(x)\)

Answer 19

now returning to \(a\) and \(b\) the right hand side is
\[\frac{a}{b}+\frac{b}{a}\left(1+\frac{1}{a^2}\right)\] with me so far?

Answer 20

@satellite73 yes!!!:D

Answer 21

now we multiply out and add the fractions

Answer 22

\[\frac{a}{b}+\frac{b}{a}\left(1+\frac{1}{a^2}\right)\]
\[=\frac{a}{b}+\frac{b}{a}+\frac{b}{a^3}\] adding the first two fractions gives
\[\frac{a^2+b^2}{ab}+\frac{b}{a^3}\]

Answer 23

now w recall that \(a^2+b^2=\cos^2(x)+\sin^2(x)=1\) so we have
\[\frac{1}{ab}+\frac{b}{a^3}\]

Answer 24

adding this fractions, the common denominator is \(a^3b\) so we get
\[\frac{a^2}{a^3b}=\frac{b^2}{a^3b}=\frac{a^2+b^2}{a^3b}\]

Answer 25

typo there sorry, it is plus, not equal \[\frac{a^2}{a^3b}+\frac{b^2}{a^3b}=\frac{a^2+b^2}{a^3b}\]

Answer 26

now once again we remember that \(a^2+b^2=1\) giving us
\[\frac{1}{a^3b}\] which is exactly what we wanted

Answer 27

@satellite73 so what would be the final answer?

Answer 28

this was a long one, but most of it was just algebra

Answer 29

well you had to show that both sides were the same, and that is what we did

Answer 30

the right hand side is
\[\frac{1}{\cos^3(x)\sin(x)}\] and so is the left hand side

Answer 31

i could try to write this out with \(\cos(x)\) and \(\sin(x)\) but it would take forever.
there were two tricks here
the first was to call \(2+\tan^2(x)=1+1+\tan^2(x)=1+\sec^2(x)\) and the rest was algebra, but recalling that \(a^2+b^2=1\)

Answer 32

i have all the steps written out without leaving any thing out
copy it down exactly, but replace \(a\) by \(\cos(x)\) and \(b\) by \(\sin(x)\) and you will have all the steps

Answer 33

for example
\[\frac{a}{b}+\frac{b}{a}\left(1+\frac{1}{a^2}\right)\] will become
\[\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}\left(1+\frac{1}{\cos^2(x)}\right)\] and so on

Answer 34

the "final answer" is really the worked out algebra, because your job was to show that the right and left hand sides were the same thing