Eliminating Theta: Steps to Follow

[sahilmm15]

Homework Statement

If $cosec\theta - sin\theta = m$ and $sec\theta - cos\theta = n$, then eliminate $\theta$

Relevant Equations

N/A

I did try a few steps:-​

After converting both the equations in terms of cos and sin I got-
$$cos^2\theta = msin\theta$$
and $$sin^2\theta = ncos\theta$$.

That's all what I am able to do. Cant find how do I eliminate $\theta$ from here

​

 

[fresh_42]

Assuming this is right what you have done, then changing to $\tan$ and $\cot$ seems to be an option.

 

[sahilmm15]

Assuming this is right what you have done, then changing to $\tan$ and $\cot$ seems to be an option.

How do we change to $tan\theta$ and $cot\theta$ to eliminate $\theta$

 

[fresh_42]

How do we change to $tan\theta$ and $cot\theta$ to eliminate $\theta$

By dividing the equations. At last you still have $\sin^2+\cos^2=1$.

 

[sahilmm15]

By dividing the equations. At last you still have $\sin^2+\cos^2=1$.

By dividing both the equations we get $cot^3\theta = {\frac{m}{n}}$

 

[fresh_42]

And? What do you get from the other hint I gave?

 

[sahilmm15]

And? What do you get from the other hint I gave?

I can't figure it out. The only way to make use of $sin^2\theta + cos^2\theta$ is by adding both the two equations??

 

[fresh_42]

I can't figure it out. The only way to make use of $sin^2\theta + cos^2\theta$ is by adding both the two equations??

I am not sure whether this is necessary. You can already solve for $\theta$ by taking the cubic root. However, the other equation can be used to check the result.

 

[sahilmm15]

I am not sure whether this is necessary. You can already solve for $\theta$ by taking the cubic root. However, the other equation can be used to check the result.

I think we need to find the independent values of $sin\theta$ and $cos\theta$ (in terms of m and n) and then put in $sin^2\theta + cos^2\theta$ to get rid of $\theta$, but that would be tedious.

 

[fresh_42]

I think we need to find the independent values of $sin\theta$ and $cos\theta$ (in terms of m and n) and then put in $sin^2\theta + cos^2\theta$ to get rid of $\theta$, but that would be tedious.

That is not what you wrote. Your text says

... then eliminate $\theta$

And under the assumption you made no mistake, you got $\theta =\operatorname{arc}\cot \sqrt[3]{\dfrac{m}{n}}$.

 

[sahilmm15]

That is not what you wrote. Your text says And under the assumption you made no mistake, you got $\theta =\operatorname{arc}\cot \sqrt[3]{\dfrac{m}{n}}$.

The question states under the given conditions eliminate $\theta$. It means there should be no $\theta$ in the answer. Then it implies answer should be in the terms of m and n( a relation between m and n where there is no $\theta$).

 

[fresh_42]

The question states under the given conditions eliminate $\theta$. It means there should be no $\theta$ in the answer. Then it implies answer should be in the terms of m and n( a relation between m and n where there is no $\theta$).

And since we have $\theta=\theta(n,m)$ we can substitute every occurrence of $\theta$ by an expression which only has $n,m$ in it. This is an elimination. And please stop shouting!

 

[sahilmm15]

And since we have $\theta=\theta(n,m)$ we can substitute every occurrence of $\theta$ by an expression which only has $n,m$ in it. This is an elimination. And please stop shouting!

To be honest, I was not shouting. And I think it is not fair of you to assume anything. There are bold and italics formatting tools to convey ones expressions and what they are thinking, and it doesn't mean one is shouting when they are using them.

 

[fresh_42]

Yes, and boldface means shouting on the internet.

I do not like that solution either, and I am not sure whether we lost information on the way by squaring or dividing. I would check it with some easy values for $n,m$.

 

[sahilmm15]

Yes, and boldface means shouting on the internet.

I do not like that solution either, and I am not sure whether we lost information on the way by squaring or dividing. I would check it with some easy values for $n,m$.

Lol, honestly I didn't knew that it meant shouting. So I misinterpreted you. I am sorry.