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Geometry Level 2

If $\tan x + \cot x = 4$ , which of the following is a possible value of $x$ ?

45^\circ

60^\circ

30^\circ

15^\circ

2 solutions

Chew-Seong Cheong
Jul 19, 2016

$\begin{aligned} \tan x + \cot x & = 4 \\ \frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} & = 4 \\ \frac {\sin^2 x + \cos^2 x}{\sin x \cos x} & = 4 \\ \frac 2{\sin 2x} & = 4 \\ \implies \sin 2x & = \frac 12 \\ 2x & = 30^\circ & \small \color{#3D99F6}{\text{For the first quadrant}} \\ \implies x & = \boxed{15^\circ} \end{aligned}$

Anurag Pandey
Jul 28, 2016

I knew tan 15°=2-√3 and cot 15°=2+√3 So adding them we get 4 .

How do you show that $\tan 15^\circ = 2-\sqrt3$ and $\cot 15^\circ = 2 + \sqrt3$ ?

Pi Han Goh - 4 years, 10 months ago

$\tan (\alpha - \beta)= \frac{\tan \alpha - \tan \beta}{1 +( \tan \alpha \times \tan \beta)}$

Now substitute \alpha =45° and

$\beta =30°$ we get the value of $\tan 15°$ .

And we know \cot 15° = \tan (90°-15°) which is equal to \tan 75° we could the formula given above just reverse the sign. (+ to - and vice versa). Substituting $\tan 30° = \frac {1}{√3}$

and \tan 45°= 1.

Sorry I should have explained it earlier.

Anurag Pandey - 4 years, 10 months ago

Thank youuuu. Another way to solve this is to apply the double angle identity , $\tan (2x) = \dfrac{2\tan x}{1 - \tan^2 x}$ , in this case, $x = 15^\circ$ .

Pi Han Goh - 4 years, 10 months ago

Ya we could use this too but in that we need to solve a quadratic and reject a root as tan 15° cannot be greater than tan 45 ° which is equal to 1 . So we reject 2 +√3 and get the answer as tan 15° = 2 - √3. 😀😀

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Great work! Another altenative solution is to apply the triple angle identity , $\tan(3x) = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$ , in this case, $x= 15^\circ$ as well.

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – Why is it that we get two roots and why do we need discard one . ??even if me made the equation properly.

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Think of it this way: You substitute $y = \tan x$ right? Is $\tan x$ a one-to-one function? What constraints do we need to consider?

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – Yup. It's a on to one function. We need the constraints that x belongs to (\ \frac {-π}{2} to \frac {π}{2} )\ for it to be one to one function.ryt.? But how does that explain that we get tan 15° = 2+√3 after we solve the quadratic ?

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Well, we know that $\tan 0^\circ < \tan 15^\circ < \tan45^\circ$ right? That's the other constraint that I'm talking about.

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – Okay I got that. But why in the first place are we getting the root as 2+√3 even though we were solving for tan15° ? Is it just that it's a quadratic so it will give two roots or anything else.?

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – That is known as an extraneous root. Basically it's "a solution to an equation that seems to be right, but when we check it (by substituting it into the original equation) turns out not to be right."

And yes, it's a quadratic equation, recall that any $n^\text{th}$ degree polynomial has exactly $n$ roots. So a quadraitc equation (a $2^\text{nd}$ degree polynomial) has exactly two roots.

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – Thank you. I will remember that. So basically even though we get a answer we need to back substitute in the original equation to check whether it satisfies it or not . Ryt?

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Yes. Exactly. Now here's an exercise that requires you to check for extraneous roots:

Find all real roots of $x^2 - 3 |x| + 10 = 0$ .

And if you want a more difficult one, try

Prove that $\tan25^\circ$ is a root of the equation $x^3 + x^2(-3\sqrt3- 6) - 3x + (\sqrt3 + 2) = 0$ .

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – For the first question I solved considering two cases for x greater than or equal to zero and another x<0. I got answer as no real roots exits. Is it correct ?

And for the second one I am still thinking .

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – For the second one. $Let 25° = \theta = \frac {5π}{36}$ Multiplying both sides by 3 we get $3 \theta = \frac {5π}{12}$ Now taking tan both sides we get $\tan 3 \theta = \tan \frac {5π}{12}$ Now we could use the formal of $\tan 3 \theta$ and then substitute $\tan \theta = x$ and $\tan \frac {5π}{12} = 2 +√3$ and just cross multiply and we would get our result . Any other way to solve it? \

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – got it! Any other way to do it? 😀😀

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – I didnt read the whole comment as I was not able to scroll sorry. I will think of something else too. Need some time.

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – There are other ways too, but this is the simplest one. Well done!

Pi Han Goh - 4 years, 10 months ago

@Anurag Pandey – Nope. Not correct. Hint : What is $\sqrt{x^2}$ ?

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – $√x^2 is equal to |x| .$

Anurag Pandey - 4 years, 10 months ago

@Pi Han Goh – But the discriminant comes out to be negative so no real solution should exist.

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Okay, here's another hint: Take $y = |x|$ .

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – I got what you are saying that we could get a quadratic in |x| but still after solving it too the answer will be that no real roots exists. If the question would have been ( x^2 - 3|x| -10 = 0 then the roots will be real and the answer would have been x= +5 or -5 . We will discard the solution that |x| =-2 as this is not possible . Ryt?

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – I got what you are saying that we could get a quadratic in |x| but still after solving it too the answer will be that no real roots exists. If the question would have been $x^2 - 3|x| -10 = 0 then the roots will be real and the answer would have been x= +5 or -5 . We will discard the solution that |x| =-2 as this is not possible . Ryt?$

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – yes correct!

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – But we could also solve the problem but considering two cases like I said before . As we will fix the domain of x if we got any solution Lying outside of hat domain we would discard it.
Ryt?

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Ryt Ryt Ryt Ryt Ryt Ryt Ryt Ryt Ryt

Pi Han Goh - 4 years, 10 months ago

@Pi Han Goh – Haha haha . 😀😀😀😀😀😀😀😀😁😁😁

Anurag Pandey - 4 years, 10 months ago

@Pi Han Goh – I am getting crazy. I was actually typing the solution and suddenly there was some network problem so I thought my comment was not posted. And to add more confusion I saw a comment @Anurag Pande so I thought you posted it. But then I realized I was my comment only. Confusion confusion!!! Anyways is the solution correct ? And is there any other way to solve it?

Anurag Pandey - 4 years, 10 months ago

You know this was my first solution in latex. I thought it might be difficult. But it was quite simple but lengthy to type and did some mistake to . But will try not to minimise these errors. . 😀😀

Anurag Pandey - 4 years, 10 months ago

Try to minimise these errors. *

Anurag Pandey - 4 years, 10 months ago

@Anurag Pandey – Haha, yeah, just keep practicing, you will get a hang of it! =D =D

Post questions too!

Pi Han Goh - 4 years, 10 months ago

I got my trigonometric table with me!

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