Eliminate "A" from equations x = 2 + cosecA and y=(1/4)tanA

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Monkeyseat

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First of all, I apologise for posting quite a few questions today. I don't like to post more than a couple of questions at once but this was an exercise of 20+ questions and there have been 3 that have stumped me so it's not like I'm posting the whole thing.

First I will show you what I have done on a similar question so the method makes sense:

Question:

Eliminate "A" from equations x = cosecA and y = (1/4)cotA

Solution:

x^2 = cosec^2 A

4y = cot A
16y^2 = cot^2 A

Therefore, as 1 + cot^2 A = cosec^2 A:

1 + 16y^2 = x^2

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Now to the similar one I can't do:

Question:

Eliminate "A" from equations x = 2 + cosecA and y = (1/4)tanA

Working:

x - 2 = cosec A
(x - 2)^2 = cosec^2 A
(x-2)^2 = 1/(sin^2 A)

4y = tanA
16y^2 = tan^2 A
16y^2 = sec^2 A - 1
16y^2 + 1 = 1/(cos^2 A)

So I have (x-2)^2 = 1/(sin^2 A) and 16y^2 - 1 = 1/(cos^2 A). However, I don't know where to go from here and what to substitute into?

I think I am getting close, the book says the answer is (x - 2)^2 = 1 + (16y^2)/(16y^2).

Any tips?

Many thanks.
 
Monkeyseat said:
Question:

Eliminate "A" from equations x = 2 + cosecA and y=(1/4)tanA

cosec A = x-2

sin A = 1/(x-2)..........................................(1)

4y = tanA

sinA = 4y/[sqrt(1+16y^2)]............................(2)

Equate (1) and (2)

Another way

cosec^2( A) = (x-2)^2 .....................(3)

cotA = 1/( 4y)

cot^2 (A) = 1/(16y^2)

cosec^2(A) = 1 + 1/(16y^2)..............(4)

Equate (3) and (4)Working:

x - 2 = cosec A
(x - 2)^2 = cosec^2 A
(x-2)^2 = 1/(sin^2 A)

4y = tanA
16y^2 = tan^2 A
16y^2 = sec^2 A - 1
16y^2 + 1 = 1/(cos^2 A)

So I have (x-2)^2 = 1/(sin^2 A) and 16y^2 - 1 = 1/(cos^2 A). However, I don't know where to go from here and what to substitute into?

I think I am getting close, the book says the answer is (x - 2)^2 = 1 + (16y^2)/(16y^2).

Any tips?

Many thanks.
Click to expand...
 
How did you get from "4y = tanA" to "cotA = 1/( 4y)"?

Putting that aside, if I equate 3 and 4:

(x - 2)^2 = 1 + (1/(16y^2))

So is that the answer? I'm just wondering becasue the book says the answer is (x - 2)^2 = 1 + (16y^2)/(16y^2). Is this wrong? Going off the book's answer, it would always equal 2????? That seems a bit weird. Could you please clarify if the book is incorrect?

Thanks.
 
Monkeyseat said:
How did you get from "4y = tanA" to "cotA = 1/( 4y)"?

cot A = 1/tan A .............. fundamental trigonometric identity

Putting that aside, if I equate 3 and 4:

(x - 2)^2 = 1 + (1/(16y^2))

So is that the answer? I'm just wondering becasue the book says the answer is

(x - 2)^2 = 1 + (16y^2)/(16y^2).



(x-2)^2 = {1 + 16y^2}/(16y^2)

whch is equivalent to

(x-2)^2 = 1/(16y^2) + 16y^2/(16y^2)

whch is equivalent to

(x-2)^2 = 1/(16y^2) + 1

whch is equivalent to

(x-2)^2 = 1 + 1/(16y^2)



Is this wrong? Going off the book's answer, it would always equal 2 (I t would be true - If they wrote it the way you wrote here - however I doubt that book wrote exactly that way)????? That seems a bit weird. Could you please clarify if the book is incorrect?

Thanks.
Click to expand...
 
Subhotosh Khan, I have a photo here of the exact answer in the back of the book:

hpim3603rm9.jpg


Is this wrong?

Thanks.
 
Monkeyseat said:
Subhotosh Khan, I have a photo here of the exact answer in the back of the book:


Is this wrong? <<< Yes - without qualification.

Thanks.
Click to expand...
 
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