Solving "tanθ + cotθ = 2": Is it Wrong?

[BOAS]

Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement

Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations

The Attempt at a Solution

tanθ + cotθ = 2

\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = 2

\frac{sin^{2}\theta}{cos\theta sin\theta} + \frac{cos^{2}\theta}{sin\theta cos\theta} = 2

\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

sin^{2}\theta + cos^{2}\theta = 2cos\theta sin\theta

Since sin^{2}\theta + cos^{2}\theta = 1

then

2cos\theta sin\theta = 1

(the above is where I think I've done something careless)

I now have

sin^{2}\theta + cos^{2}\theta = 1

I don't know what to do with it really... I tried saying sin^{2}\theta = 1 - cos^{2}\theta but it doesn't really get me anywhere.Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.

 

[SteamKing]

Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.

 

[BOAS]

Well, manipulating the original equation gives 2cosθsinθ = 1. What you need to do is find θ to make this equation true.

Ahhh, I was focused on the other side of the equation!

2cosθsinθ = sin2θ

sin2θ = 1

2θ = 90

θ = 45

from the curve I can see where -135 comes from also.

 

[Mark44]

It's quicker to replace cot(θ) by 1/tan(θ)

tanθ + cotθ = 2
tanθ + 1/tanθ = 2

Multiply by tanθ to get
tan²θ + 1 = 2tanθ
tan²θ - 2tanθ + 1 = 0
This equation can be factored to solve for tanθ.

 

[Ray Vickson]

Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement

Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations

The Attempt at a Solution

tanθ + cotθ = 2

\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = 2

\frac{sin^{2}\theta}{cos\theta sin\theta} + \frac{cos^{2}\theta}{sin\theta cos\theta} = 2

\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

sin^{2}\theta + cos^{2}\theta = 2cos\theta sin\theta

Since sin^{2}\theta + cos^{2}\theta = 1

then

2cos\theta sin\theta = 1

(the above is where I think I've done something careless)

I now have

sin^{2}\theta + cos^{2}\theta = 1

I don't know what to do with it really... I tried saying sin^{2}\theta = 1 - cos^{2}\theta but it doesn't really get me anywhere.


Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.

The quantity $t = \tan(\theta)$ must satisfy the equation
t + \frac{1}{t} = 2,
whose unique (real) solution is $t = 1$. This is easy to see and prove. For any $A,B > 0$ the arithmetic-geometric inequality says
\frac{A+B}{2} \geq \sqrt{AB},
with equality only when $A = B$. Apply this to $A = t, B = 1/t,$ to conclude that for $t > 0$ we have $t + 1/t \geq 2,$ with equality only when $t = 1$. For $t < 0$ it similarly follows that $t + 1/t \leq -2,$ so negative values of $t$ won't work.

So, the unique root is $\tan(\theta) = 1$.

 

[BOAS]

It's quicker to replace cot(θ) by 1/tan(θ)

tanθ + cotθ = 2
tanθ + 1/tanθ = 2

Multiply by tanθ to get
tan²θ + 1 = 2tanθ
tan²θ - 2tanθ + 1 = 0
This equation can be factored to solve for tanθ.

This is how I solved it after struggling with the method outlined in the OP.

Much simpler, but it is cool to see the many ways of solving these problems.

 

[Anonymoose2]

Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Homework Statement

Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

Homework Equations

The Attempt at a Solution

tanθ + cotθ = 2

\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = 2

\frac{sin^{2}\theta}{cos\theta sin\theta} + \frac{cos^{2}\theta}{sin\theta cos\theta} = 2

\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

sin^{2}\theta + cos^{2}\theta = 2cos\theta sin\theta

Since sin^{2}\theta + cos^{2}\theta = 1

then

2cos\theta sin\theta = 1

(the above is where I think I've done something careless)

Yes, you've said that \frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

and sin^{2}\theta + cos^{2}\theta = 2cos\theta sin\theta

Now, substitute the 2cos\theta sin\theta into the numerator of \frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

This will give you \frac{2cos\theta sin\theta}{cos\theta sin\theta} = 2

which simplifies into 2 = 2

 

[ehild]

2cosθsinθ = sin2θ

sin2θ = 1

2θ = 90

θ = 45

from the curve I can see where -135 comes from also.

sin(2θ)= 1 when 2θ =90 °+ k 360 °, k=0, ±1 ±2... . That means θ=45 ° ± k 180 °. Which angles are in the range [-180°, 180°]?

ehild

 

[Mark44]

Yes, you've said that \frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

and sin^{2}\theta + cos^{2}\theta = 2cos\theta sin\theta

Now, substitute the 2cos\theta sin\theta into the numerator of \frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta} = 2

This will give you \frac{2cos\theta sin\theta}{cos\theta sin\theta} = 2

which simplifies into 2 = 2

So? The idea is to show that tanθ + cotθ = 2, not that 2 = 2, which is obviously true.