How Do You Solve This Complex Trigonometry Homework Problem?

[Hockeystar]

Homework Statement

if tan(x) + cot(x) = 4 then what is

cos^2(x) + sin^2(x) + tan^2(x) + cot^2(x) + csc^2(x) + sec^2(x)

Homework Equations

cos^2(x) + sin^2(x) = 1
cot^2(x) = 1/(tan^2(x))

The Attempt at a Solution

= 1 + 14 + 1/(cos^2(x)sin^2(x))

Now I'm stuck.

 

[Mark44]

Homework Statement

if tan(x) + cot(x) = 4 then what is

cos^2(x) + sin^2(x) + tan^2(x) + cot^2(x) + csc^2(x) + sec^2(x)

Homework Equations

cos^2(x) + sin^2(x) = 1
cot^2(x) = 1/(tan^2(x))

The Attempt at a Solution

= 1 + 14 + 1/(cos^2(x)sin^2(x))

Now I'm stuck.

How did you get this? "= 1 + 14 + 1/(cos^2(x)sin^2(x))"
What's the other side of this equation?

 

[Hockeystar]

square first equation and you get tan^2(x) + cot^2(x) +2 = 16

 

[Mark44]

You don't "square an equation;" you can square each side of an equation. Are you trying to say that (tan (x) + cot(x))² = tan²(x) + cot²(x)?

How about (3 + 4)²? Is that equal to 3² + 4²?

 

[TheoMcCloskey]

Can we not solve for the first equation for $x$?

Express first equation in terms of $\sin(x)$ and $\cos(x)$, get common denominator, and solve for $x$.

Second equation can be simplified some-what by noting some additional trig identies for $\csc^2(x)$ and $\sec^2(x)$, but we can use solution for $x$ from first equation and substitute to solve for second equation directly.

 

[Mark44]

Can we not solve for the first equation for $x$?

Express first equation in terms of $\sin(x)$ and $\cos(x)$, get common denominator, and solve for $x$.

Second equation can be simplified some-what by noting some additional trig identies for $\csc^2(x)$ and $\sec^2(x)$, but we can use solution for $x$ from first equation and substitute to solve for second equation directly.

There is no second equation. The idea is to simply evaluate that expression using the value(s) of x obtained from the first (and only) equation.

The first equation can be solved with minimal substitution, keeping things in terms of tan(x).

 

[vela]

You don't "square an equation;" you can square each side of an equation. Are you trying to say that (tan (x) + cot(x))² = tan²(x) + cot²(x)?

No, he's saying

$$(\tan x+\cot x)^2=\tan^2 x+2 \tan x\cot x +\cot^2 x=\tan^2 x+2+\cot^2 x$$

since tan x cot x=1. Because the LHS is also equal to 4², he finds

$$\tan^2 x+\cot^2 x=16-2=14$$.

 

[vela]

tan(x) + cot(x) = 4

Write this in terms of sin x and cos x and put the LHS over a common denominator. It'll simplify a bit, giving you a result that'll let you finish the problem.

 

[Mentallic]

Homework Statement

if tan(x) + cot(x) = 4 then what is

cos^2(x) + sin^2(x) + tan^2(x) + cot^2(x) + csc^2(x) + sec^2(x)

Homework Equations

cos^2(x) + sin^2(x) = 1
cot^2(x) = 1/(tan^2(x))

The Attempt at a Solution

= 1 + 14 + 1/(cos^2(x)sin^2(x))

Now I'm stuck.

Take a look back at the original equality $tanx+cotx=4$ and note that $tanx=sinx/cosx$ and $cotx=cosx/sinx$

 

[Hockeystar]

just solved it and I was so close. 1/(cos(x)sin(x)) = 4 Substitute that in for 1/(cos^2(x)sin^2(x)). You get 16. So the answer is 1 + 14 + 16 = 31

 

[Mentallic]

Yeah, you were close

 

[ƒ(x)]

You don't "square an equation;" you can square each side of an equation. Are you trying to say that (tan (x) + cot(x))² = tan²(x) + cot²(x)?

How about (3 + 4)²? Is that equal to 3² + 4²?

Lets not be hasty to bash other's ideas. The cot and tan cancel out, leaving 2.

 

[Mark44]

I misread that final +2 on the left side in what Hockeystar posted, probably misreading it as the exponent on cot(x). My mistake.
"tan^2(x) + cot^2(x) +2 = 16"

 

[snshusat161]

Given: tan(x) + cot (x) = 4

Square both side then,

$$tan^{2}x + cot^{2}x$$ + 2.tan(x) cot(x) = 16

but tan(x). cot(x) = 1

therefore, $$tan^2{x} + cot^2{x}$$ = 14.

Take this equation first.

 

[snshusat161]

Now again consider tan(x) + cot(x) = 4

Now tan(x) = $$\frac{sin(x)}{cos(x)}$$ and similarly cot(x) = $$\frac{cos(x)}{sin(x)}$$

so

$$\frac{sin(x)}{cos(x)}$$ + $$\frac{cos(x)}{sin(x)}$$ = 4

$$\frac{sin^{2}x + cos^{2}x}{sin(x).cos(x)}$$ = 4

then it is equal to

$$\frac{1}{sin(x).cos(x)} = 4$$

take it equation second.

 

[snshusat161]

Now take your actual equation

$$cos^{2}(x) + sin^{2}(x) + tan^{2}(x) + cot^{2}(x) + csc^{2}(x) + sec^{2}(x)$$

Now here, $$cos^{2}(x) + sin^{2}(x)$$ = 1

$$tan^{2}(x) + cot^{2}(x)$$ = 14

 

[snshusat161]

now last thing left is $$csc^{2}(x) + sec^{2}(x)$$

$$\frac{1}{sin^{2}x}$$ + $$\frac{1}{cos^{2}x}$$

that is equal to

$$\frac{cos^{2}x + sin^{2}x}{sin^2(x). cos^{2}x}$$

now look at equal second and I should leave rest up to you so that you can feel that you have done your home not me.

 

[snshusat161]

now last thing left is $$csc^{2}(x) + sec^{2}(x)$$

$$\frac{1}{sin^{2}x}$$ + $$\frac{1}{cos^{2}x}$$

that is equal to

$$\frac{cos^{2}x + sin^{2}x}{sin^2(x). cos^{2}x}$$

now look at equal second and I should leave rest up to you so that you can feel that you have done your home not me.

 

[Mentallic]

Given: tan(x) + cot (x) = 4

Square both side then,

$$tan^{2}x + cot^{2}x$$ + 2.tan(x) cot(x) = 16

but tan(x). cot(x) = 1

therefore, $$tan^2{x} + cot^2{x}$$ = 14.

Take this equation first.

...
Now again consider tan(x) + cot(x) = 4

Now tan(x) = $$\frac{sin(x)}{cos(x)}$$ and similarly cot(x) = $$\frac{cos(x)}{sin(x)}$$

so

$$\frac{sin(x)}{cos(x)}$$ + $$\frac{cos(x)}{sin(x)}$$ = 4

$$\frac{sin^{2}x + cos^{2}x}{sin(x).cos(x)}$$ = 4

then it is equal to

$$\frac{1}{sin(x).cos(x)} = 4$$

take it equation second.

...
Now take your actual equation

$$cos^{2}(x) + sin^{2}(x) + tan^{2}(x) + cot^{2}(x) + csc^{2}(x) + sec^{2}(x)$$

Now here, $$cos^{2}(x) + sin^{2}(x)$$ = 1

$$tan^{2}(x) + cot^{2}(x)$$ = 14

.....
now last thing left is $$csc^{2}(x) + sec^{2}(x)$$

$$\frac{1}{sin^{2}x}$$ + $$\frac{1}{cos^{2}x}$$

that is equal to

$$\frac{cos^{2}x + sin^{2}x}{sin^2(x). cos^{2}x}$$

now look at equal second and I should leave rest up to you so that you can feel that you have done your home not me.

Oh wow, you gave such a subtle hint there... the OP would definitely feel like the homework was completed entirely on his/her own...

By the way, the OP had all except the $tan^2x+cot^2x$ part figured out before even creating this thread and then did figure out the answer entirely shortly after. You would have known this if you read through the thread.

 

[snshusat161]

Oh wow, you gave such a subtle hint there... the OP would definitely feel like the homework was completed entirely on his/her own...

By the way, the OP had all except the $tan^2x+cot^2x$ part figured out before even creating this thread and then did figure out the answer entirely shortly after. You would have known this if you read through the thread.

Oh! I have seen only the last post and thought the question is still unsolved and so I've given the solution, but I think Mark got an answer that about the last step. Secondly I've not completed this question entirely cause I've seen somewhere in this forum that we have to not give complete solution for any problem. Is I'm right?

 

[Mentallic]

Secondly I've not completed this question entirely cause I've seen somewhere in this forum that we have to not give complete solution for any problem. Is I'm right?

Well you basically left a complete solution by stopping near the end where the rest requires no more thinking on the OP's behalf.

It's like saying and then y=1+14+16 but I won't finish simplifying that because I can't provide a complete solution. You already have.

But no harm done, the OP already had the answer.

 

[snshusat161]

no, there is an answer in the second equation. He has to only square and add. look

 

[snshusat161]

$$\frac{1}{sin(x).cos(x)} = 4$$

This is what I've given in the second equation.

 

[snshusat161]

$$\frac{1}{sin(x).cos(x)} = 4$$

This is what I've given in the second equation.

 

[snshusat161]

how much simpler I should go to prevent myself running agains PF rule. I think I should tell him them do $$sin^{2}x + cos^{2}x$$ = 1 and then you will be left with $$\frac{1}{sin^{2}(x).cos^{2}(x)}$$ which is equal to $$4^{2}$$

 

[vela]

Look at how others responded in this thread and in other threads to see what is appropriate and what isn't.

 

[snshusat161]

@vela. I know but mentallic is directly/indirectly indicating that I've left solving it cause I don't know the solution after that part.

 

[Mentallic]

@vela. I know but mentallic is directly/indirectly indicating that I've left solving it cause I don't know the solution after that part.

I just can't wrap my head around what you mean here...?

 

[Mentallic]

no, there is an answer in the second equation. He has to only square and add. look

Exactly what I was saying. You've left nothing to the imagination which is why you've given the solution indirectly and this makes your fear of not giving the entire solution to abide by PF rules futile.

 

[snshusat161]

I just can't wrap my head around what you mean here...?

I mean you want to show that I'm not following PF rules

 

[snshusat161]

Exactly what I was saying. You've left nothing to the imagination which is why you've given the solution indirectly and this makes your fear of not giving the entire solution to abide by PF rules futile.

I think trigonometry is very difficult part for noobs. That's the only reason.

 

[Mentallic]

I mean you want to show that I'm not following PF rules

I didn't mean to provoke you, I was trying to show you why you weren't actually obeying the PF rules. If you disagree with me, then by all means, ask a moderator if your method of helping is within the guidelines of the rules.

I think trigonometry is very difficult part for noobs. That's the only reason.

This doesn't mean we should be solving all the trigonometry and leaving just the algebra part up to them to solve, because they're not as "noob" at algebra.
Also, the OP isn't as bad at trigonometry as you make him out to be And we've all been there at one stage in our lives. Believe it or not, I too was a noob at trigonometry about 2 years back.

 

[snshusat161]

Okay, I'll take care next time I post.

 

[Mentallic]

I myself had an infraction when I started off in this forum so don't worry, it'll quickly become clear to you what is acceptable and what isn't. Most if not all high-count poster (I'll say 500 and above) give acceptable homework help. You should follow their lead