Question

sec^6x(secxtanx)-sec^4x(secxtanx)=sec^5xtan^3x verify

Answer 1

i have the method but have you tried this?

Answer 2

tried what?

Answer 3

if we look at the LHS we have secxtanx present in both terms, so this can be factored out

comes to (sec^6x - sec^4x)(secxtanx)

what do you think the next step will be?

Answer 4

oh well it becomes sec^2x which becomes 1+tan^2x

Answer 5

actually nevermind that doesn't work I don't know what to do next

Answer 6

next step is

(sec^4x.sec^2x - sec^4x)(secx.tanx)

you follow so far

Answer 7

so you just broke apart the sec^6x

Answer 8

yep you got it, now for the next step

[sec^4x(sec^2x -1)](secxtanx)

happy with that?

Answer 9

yeah

Answer 10

thank you! can I ask another

Answer 11

we know that sec^2x = 1+tan^2x

so we have (sec^4x.tan^2x)(secxtanx)

what do you think we do now?

Answer 12

multiply and your done

Answer 13

that's it!! and it equals the RHS

Answer 14

six(1-2cos^2x+cos^4x)=sin^5x

Answer 15

ok give me a few minutes

Answer 16

got it

Answer 17

whats the six doing right at the front?

Answer 18

there is no six?

Answer 19

This is what you wrote

six(1-2cos^2x+cos^4x)=sin^5x

Answer 20

yeah I don't see a six?

Answer 21

oh the word six its sinx

Answer 22

that's what i thought it might be

Answer 23

I factored the cos^4x and got no where

Answer 24

what did you factor it to?

Answer 25

(1-cos^2x)^2

Answer 26

oh i see

should be (cos^2x -1)^2

Answer 27

got it that doesn't simplify tho to something else

Answer 28

yes it does!

we know that cos^2x + sin^2x =1

so what do you think that cos^2x - 1 will equal?

Answer 29

-sin^2x but how does that equal positive sin on the other side

Answer 30

oh its squared I got it

Answer 31

well done

Answer 32

it becomes sin^4x.sinx which is equal to the RHS

Answer 33

yeah

Answer 34

4tan^4x+tan^2x-3=sec^2x(4tan^2x-3)

Answer 35

ok, let's do this together, give me a min or two

Answer 36

I have this and one more then im done

Answer 37

OK standby

Answer 38

this time we look at the RHS....

we know that sec^2x = tan^2x + 1

so we have... (tan^2x +1)(4tan^2x - 3)

now what's next?

Answer 39

really you just multiply and that's it?

Answer 40

why couldn't I notice that before

Answer 41

because these problems require practice, a keen eye and lots of thinking.
i think you're on your way to achieving all three

Answer 42

multiply this out and you get the LHS

Answer 43

(1+sec(-x))/(sin(-x)+tan(-x))=-cscx

Answer 44

ok, standby

Answer 45

yes

Answer 46

ok here we go, this one is a bit more tricky......

we know that sec(-x) = secx and sin(-x) = -sinx and tan(-x) = -tanx
so looking at the LHS we now have.........

(1 + secx)/-(sinx +tanx)

happy with that? what do you think we do now?

Answer 47

why does the negative for sinx go outside the( )

Answer 48

because sin(-x) + tan(-x) = -sinx - tanx = -(sinx + tanx)

Answer 49

this is true yes

Answer 50

ok, got any ideas for the next step?

Answer 51

this is where I have no idea

Answer 52

ok well whats another way of expressing secx? and tanx?

Answer 53

1/cosx sinx/cosx

Answer 54

like I know you can swap those but then it gets ugly

Answer 55

yes you're right and another way of expressing tanx is sinx.secx
you ok with that?

Answer 56

yeah

Answer 57

ok so we substitute sinx.secx for tan x in the bottom line and we get

(1 + secx)/-(sinx + sinx.secx)

can you see where i'm going?

Answer 58

no it still looks like I have no idea where to go after that

Answer 59

ok look at it carefully

the bottom line can now be factored out with sinx

can you see?

Answer 60

well yeah but how does that help with the top

Answer 61

because then you'll have (1 + secx) on the top and bottom which cancels out!

Answer 62

I see I was like where are you going with this...but see I would never realize that on my own

Answer 63

so we're left with -1/sinx which is equal to the RHS

Answer 64

if you look at things hard enough and do a bit of trial and error with heaps of logical thinking then you can do this too!

Answer 65

always use the basic trig identities to work these questions out

Answer 66

was that the last one?

Answer 67

yes so your saying when you can switch from cos^2x to 1-sin^2x to always try and do that?

Answer 68

yes that's one of the many trig identities, there's heaps more.
just look at your text books or the internet, but that's the basic idea.
trial and error until you work it out, then you"ll just get better at it

Answer 69

see you next time