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Trig equation- my error, or the textbook?

1+cotx1+tanx=5 \frac{1+cot x}{1+tan x}=5

I don't know LaTex well but the interval is 0(less than or equal to)x(less than or equal to) 360 degrees.

I derived the solutions 11.3, 135, 191, 315 (3 s.f) However, the textbook solution only has two of these, 11.3 and 191. My hypothesis is that the textbook solution only used one of two possible values of tan theta from the quadratic, while I used both. I may be wrong on this.

I'd attach screenshots to explain but my phone is not being particularly responsive right now.

Did I make the mistake, or is my textbook incorrect?

Thanks :smile:
Original post by Illidan2
1+cotx1+tanx=5 \frac{1+cot x}{1+tan x}=5

I don't know LaTex well but the interval is 0(less than or equal to)x(less than or equal to) 360 degrees.

I derived the solutions 11.3, 135, 191, 315 (3 s.f) However, the textbook solution only has two of these, 11.3 and 191. My hypothesis is that the textbook solution only used one of two possible values of tan theta from the quadratic, while I used both. I may be wrong on this.

I'd attach screenshots to explain but my phone is not being particularly responsive right now.

Did I make the mistake, or is my textbook incorrect?

Thanks :smile:


The issue with 135 and 315 is that then 1+tan⁡x=01+\tan x = 0 and we cannot have this because that's in the denominator of the equation.

It's good practice to note small things like these down before starting a question.
Original post by Illidan2
1+cotx1+tanx=5 \frac{1+cot x}{1+tan x}=5

I don't know LaTex well but the interval is 0(less than or equal to)x(less than or equal to) 360 degrees.

I derived the solutions 11.3, 135, 191, 315 (3 s.f) However, the textbook solution only has two of these, 11.3 and 191. My hypothesis is that the textbook solution only used one of two possible values of tan theta from the quadratic, while I used both. I may be wrong on this.

I'd attach screenshots to explain but my phone is not being particularly responsive right now.

Did I make the mistake, or is my textbook incorrect?

Thanks :smile:


have you checked your answers by putting them back into the equation?
Reply 3
Avatar for Illidan2
Illidan2
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I have since, and I can see that by feeding them back in, 135 and 315 don't work. However, because I don't know all the exact trig values, i'd have never noticed that the denominator would equal zero for these values. What is the best way to learn exact trig values?

Original post by isiaiah d
have you checked your answers by putting them back into the equation?
Original post by Illidan2
I have since, and I can see that by feeding them back in, 135 and 315 don't work. However, because I don't know all the exact trig values, i'd have never noticed that the denominator would equal zero for these values. What is the best way to learn exact trig values?


I never learnt them (apart from like right before my entrance exams - but I had forgotten them a month later).I just plugged my answers back into the question after every trig question
Reply 5
Avatar for Illidan2
Illidan2
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Original post by isiaiah d
I never learnt them (apart from like right before my entrance exams - but I had forgotten them a month later).I just plugged my answers back into the question after every trig question


A useful tactic- i'll do my best to get into the habit of employing it. Thank you! :smile:
Reply 6
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Illidan2
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Original post by RDKGames
The issue with 135 and 315 is that then 1+tan⁡x=01+\tan x = 0 and we cannot have this because that's in the denominator of the equation.

It's good practice to note small things like these down before starting a question.


Thank you :smile:
Your method is incorrect if it's throwing up those values as solutions whether you noticed 1 + tan(x)=0 or not. How are you simplifying the original expression?
Reply 8
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Illidan2
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Original post by SerBronn
Your method is incorrect if it's throwing up those values as solutions whether you noticed 1 + tan(x)=0 or not. How are you simplifying the original expression?


I got a quadratic expression in tan x, which I in turn solved. This yielded (5 tan x-1)(tan x+1)=0. Now I can see that I wasn't supposed to use tan x+1=0, as tan x cannot be -1. I then proceeded to use a CAST diagram using both of these values over the given period, when I should have only used one. Other than this, though, what about my method is incorrect?
Original post by Illidan2
I got a quadratic expression in tan x, which I in turn solved. This yielded (5 tan x-1)(tan x+1)=0. Now I can see that I wasn't supposed to use tan x+1=0, as tan x cannot be -1. I then proceeded to use a CAST diagram using both of these values over the given period, when I should have only used one. Other than this, though, what about my method is incorrect?
I'm guessing you started off by multiplying through by (tan x + 1), which is obviously going to introduce false solutions when tan x + 1 = 0.
There's no need for quadratics ... just simplify it:

1+cot(x)1+tan(x)=sin(x)+cos(x)sin(x)sin(x)+cos(x)cos(x)=cos(x)sin(x)=cot(x)=5 {1 + cot(x) \over 1 + tan(x) } = { {sin(x) + cos(x) \over sin(x)} \over {sin(x) + cos(x) \over cos(x)}} = {cos(x) \over sin(x)} = cot(x) = 5
(edited 5 years ago)
Original post by SerBronn
There's no need for quadratics ... just simplify it:

1+cot(x)1+tan(x)=sin(x)+cos(x)/oversin(x)sin(x)+cos(x)cos(x)=cos(x)sin(x)=cot(x)=5 {1 + cot(x) \over 1 + tan(x) } = { {sin(x) + cos(x) /over sin(x)} \over {sin(x) + cos(x) \over cos(x)}} = {cos(x) \over sin(x)} = cot(x) = 5


A sleeker method doesn't mean his longer one is incorrect :smile:

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