OpenStudy (anonymous):

I need some help solving this identity: cot x sec^4x = cot x + 2 tan x + tan^3x

OpenStudy (unklerhaukus):

what do you mean ?

OpenStudy (anonymous):

"Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation." These are the exact directions. I guess I have to make the right look the like the left.

OpenStudy (unklerhaukus):

first convert every term to sines and cosines \[\tan\theta=\tfrac{\sin x}{\cos x}\]\[\csc\theta=\tfrac1{\sin\theta}\]\[\sec\theta=\tfrac1{\cos\theta}\]\[\cot\theta=\tfrac1{\tan\theta}=\tfrac{\cos\theta}{\sin\theta }\]

OpenStudy (anonymous):

I did that and that's where I got lost and/or made a mistake.

OpenStudy (unklerhaukus):

what did you get for the left hand side?

OpenStudy (anonymous):

(1/tanx) + (1+tan2x)(sec2x)

OpenStudy (unklerhaukus):

huh ?

OpenStudy (anonymous):

that's what I got, so now you see my problem.

OpenStudy (unklerhaukus):

the left hand side cot x * sec^4x becomes cos x /sin x * 1/ cos^4 x simplify this further

OpenStudy (anonymous):

wouldn't it then become: cosx/sinx * 1/(cos^2x)^2 then that would become cosx/sinx * 1/(1-sin^2)^2 ??

OpenStudy (unklerhaukus):

why would you do that ?

OpenStudy (anonymous):

I have no idea what I'm doing.

OpenStudy (anonymous):

Can you start from the beginning and show me step by step?

OpenStudy (unklerhaukus):

cot x sec^4x [cot = 1/ tan = cos / sin] [sec = 1/ cos] = cos x /sin x * 1/ cos^4 x

OpenStudy (anonymous):

okay now what about the other side? How do I make them match?

OpenStudy (unklerhaukus):

first simplify cos x /sin x * 1/ cos^4 x a little

OpenStudy (unklerhaukus):

[one of the cosines cancel]

OpenStudy (anonymous):

would it be sinx * 1/cos^3x

OpenStudy (unklerhaukus):

1/ sin x * cos^3 x

OpenStudy (unklerhaukus):

Now look at the right hand side of the equation cot x + 2 tan x + tan^3x what is this in terms of sines and cosines ?

OpenStudy (anonymous):

Okay... cosx/sinx + 2(sinx/cosx) and I don't know how to simplify tan^3x

OpenStudy (unklerhaukus):

tan^3 x = sin^3 x / cos^3 x

OpenStudy (unklerhaukus):

so the right hand side is cosx/sinx + 2(sinx/cosx) + sin^3 x / cos^3 x

OpenStudy (unklerhaukus):

now compare the left with the right 1/ sin x * 1/ cos^3 x = cosx/sinx + 2(sinx/cosx) + sin^3 x / cos^3 x

OpenStudy (unklerhaukus):

multiply both sides of the equation by sin x

OpenStudy (anonymous):

is it: sin^2x * sinx/cos^3x = sinxcosx/sinx + 2(sin^2x/cosx) + sin^4x/cos^3x ??

OpenStudy (unklerhaukus):

almost, the sine cancels on the left to get 1/cos^3 x = sin x * cos x/sin x + 2(sin^2 x/cos x) + sin^4 x / cos^3 x the sine also cancels on the first term on the right hand side

OpenStudy (anonymous):

how does the sin^2x and sinx on the right cancel? Wouldn't I still have a sinx left?

OpenStudy (unklerhaukus):

i dont know where you got the sin^2x from

OpenStudy (anonymous):

I meant on the left side?

OpenStudy (anonymous):

Okay so I now have cos^3x = cosx + 2(sin^2 x/cos x) + sin^4 x / cos^3 x ?

OpenStudy (unklerhaukus):

the term on the left should be 1/cos^3x

OpenStudy (unklerhaukus):

so to make the left hand side simplify to 1 , multiply both sides of the equation by cos^3 x

OpenStudy (anonymous):

1= cos^4x + 2(cos^3xsin^2x / cosx) + cos^3sin^4x/cos^3) then 1= cos^4x + 2 (cos^2xsin^2x) + sin^4x

OpenStudy (anonymous):

is that right?

OpenStudy (unklerhaukus):

yes that's right good work, so now look at this left hand side cos^4x + 2 (cos^2xsin^2x) + sin^4x this can be simplified by factorising to make this easier to see you could temporality sub A = cos^2 x, and B = sin^2 x which gives A^2 + 2AB + B^2 can you simplify this?