MHB Solving Identifying $\theta$ in Trigonometric Equation

[paulmdrdo1]

Kindly help me with this problem. I'm stuck!

$\frac{\csc\theta+1}{\cot\theta}=\frac{\cot\theta}{\csc\theta-1}$

this is how far I can get to

$\sec\theta+\tan\theta=\frac{1}{\sec\theta-\tan\theta}$

 

[Ackbach]

Kindly help me with this problem. I'm stuck!

$\frac{\csc\theta+1}{\cot\theta}=\frac{\cot\theta}{\csc\theta-1}$

this is how far I can get to

$\sec\theta+\tan\theta=\frac{1}{\sec\theta-\tan\theta}$

I would start with
$$\sin^2(\theta)+\cos^2(\theta)=1,$$
divide through by $\sin^2(\theta)$, and see what happens.

 

[paulmdrdo1]

I would start with
$$\sin^2(\theta)+\cos^2(\theta)=1,$$
divide through by $\sin^2(\theta)$, and see what happens.

I would have this fundamental identity

$1+\cot^{2}\theta=\csc^{2}\theta$ where should I put this?

 

[Ackbach]

I would have this fundamental identity

$1+\cot^{2}\theta=\csc^{2}\theta$ where should I put this?

Right. So when I see the $\csc(\theta)-1$ on one side of the equation (I mean the identity you're trying to prove), and a $\csc(\theta)+1$ on the other side of the equation, my thoughts immediately go to the Difference of Two Squares factoring pattern. Can you see what might need to happen here?

 

[paulmdrdo1]

Right. So when I see the $\csc(\theta)-1$ on one side of the equation (I mean the identity you're trying to prove), and a $\csc(\theta)+1$ on the other side of the equation, my thoughts immediately go to the Difference of Two Squares factoring pattern. Can you see what might need to happen here?

Here it is

Using this identity $\cot^{2}\theta=(\csc\theta+1)(\csc\theta-1)$

$(\csc\theta+1)=\frac{\cot^{2}\theta}{\csc\theta-1}$

$\frac{\cot^{2}\theta}{\cot(\csc\theta-1)}=\frac{\cot\theta}{\csc\theta-1}$

$\frac{\cot\theta}{\csc\theta-1}=\frac{\cot\theta}{\csc\theta-1}$

 

[Ackbach]

Here it is

Using this identity $\cot^{2}\theta=(\csc\theta+1)(\csc\theta-1)$

$(\csc\theta+1)=\frac{\cot^{2}\theta}{\csc\theta-1}$

$\frac{\cot^{2}\theta}{\cot(\csc\theta-1)}=\frac{\cot\theta}{\csc\theta-1}$

You've the basic idea right, but you don't want to write $\cot(\csc(\theta)-1)$. Instead, write $\cot(\theta)(\csc(\theta)-1)$.

$\frac{\cot\theta}{\csc\theta-1}=\frac{\cot\theta}{\csc\theta-1}$

You need to clean up your equations - there are some typos in there - but I think you've got the basic idea.

 

[paulmdrdo1]

You've the basic idea right, but you don't want to write $\cot(\csc(\theta)-1)$. Instead, write $\cot(\theta)(\csc(\theta)-1)$.
You need to clean up your equations - there are some typos in there - but I think you've got the basic idea.

I'm using my phone right now. That is why my typing is messed up. Thanks for your help though.

 

[Ackbach]

Thanks for your help though.

You're very welcome!