Question

Please help me with this problem! I have a bit of trouble with it.

1+sec x/tan x + sin x = csc x

I get stuck halfway through.

Answer 1

is that (1+secx)/tanx......

Answer 2

It's (1 + secx)/(tanx+sinx)

Answer 3

I have this written down so far.

Answer 4

Try turning it into (1+secx)*(cotx+cscx) and foiling it out

Answer 5

I don't get a square at the end. I end up getting cotx+cscx+secxcotx+cscxsecx

Answer 6

unless you want me to simplify it?

Answer 7

which cancels to cotx+cscx+cscx+1/sinxcosx

Answer 8

cotx+2cscx+2/sin2x

Answer 9

which dosent work... hymm let me try something else

Answer 10

i know some cosines are supposed to cross out but i'm not very sure after that.

Answer 11

unless the 1 should be converted to cosx/cosx?

Answer 12

did you try dividing the right hand side into the left, leaving (....)=1 i know the tanx will cancel then

Answer 13

then you get 1+secx/1+cosx = 1

Answer 14

but the answer is cscx? hmmm

Answer 15

well, what you have is an equation, i would steer clear of thinking one side is the "answer" of the other, it is algebraically allowed to divide the LHS (left hand side) by the RHS leaving (1+secx)/(1+cosx)=cscx/cscx

Answer 16

ah that's true

Answer 17

OOPS, i made a mistake

Answer 18

i tried to solve it again and i get this so far

Answer 19

ok so what you want to do is multiply the right hand side by (sinx+tanx)/(sinx+tanx)

Answer 20

if you do that then you will see that they are equal

Answer 21

wait the bottom can be factored...

Answer 22

because sinxcosx+sinx also equals sinx(1+cosx)

Answer 23

\[\frac{ 1+\sec x }{ \tan x+\sin x }=\csc x * (\frac{ \sin x+tanx }{ \sin x+tanx })\]

Answer 24

^^^ this is correct

Answer 25

let me try it out

Answer 26

do you see what i did, the RHS is being multiplied by one

Answer 27

rhs? sorry i'm not sure what that is

Answer 28

Right hand side, the side that has the cscx term

Answer 29

OH okay

Answer 30

a question i wish i would have asked my self sooner on this problem is "can i multiply one of the sides by 1 (in some form) to make it equal the other"

Answer 31

oh yeah, that's right

Answer 32

i forgot about that

Answer 33

so did i lol

Answer 34

Here is what I tell my students to do. Look at the more complex side and turn everything into terms of sines and cosines.\[\frac{ 1+\sec \theta }{ \tan \theta+\sin \theta }=\frac{ 1+\frac{ 1 }{ \cos \theta } }{ \frac{ \sin \theta }{ \cos \theta } +\sin \theta}\]

Then get a common denominator on top and on bottom\[=\frac{ \cos \theta+1 }{ \cos \theta } \div \frac{ \sin \theta+\sin \theta \cos \theta }{ \cos \theta }\]

Then multiply by reciprocal and factor a sine theta\[=\frac{ \cos \theta+1 }{ \cos \theta } \times \frac{ \cos \theta }{ \sin \theta \left( 1+\cos \theta \right) }\]

\[=\frac{ 1 }{ \sin \theta } \]
\[=\csc \theta \]

Answer 35

@goberds and @Edutopia

Answer 36

I dont understand what you did on the "common denominator step"

Answer 37

did you multiply both sides by the reciprocal of the RHS?

Answer 38

1+1/cosx =(cosx+1)/cosx and

sinx/cosx +sinx = (sinx +sinxcosx)/cosx

Answer 39

\[1+\frac{ 1 }{ \cos \theta}=\frac{ \cos \theta }{ \cos \theta }*1+\frac{ 1 }{ \cos \theta }\]
equals

\[\frac{ \cos \theta+1 }{ \cos \theta } \]

Answer 40

Then I did the same thing for the denominator.