But will $(\tan(x))^{\ln(\sin(x))}$ ever equal $(\ln(\cot(x)))^{\cos(x)}$

Question

So I was looking through the homepage of Youtube, looking for more math problems that I thought that I might be able to solve when I got bored, since either I had already solved them before, I had attempted them and given up or they already had a question asked about them (which I found on Approach $0$), so I decided to look through my profile and look for any problems that I thought would be good solving them, but with a twist when I saw this post of mine here, which made me think

I know that $$(\tan(x))^{\sin(x)}$$ is equal to $$(\cot(x))^{\cos(x)}$$ when $x$ is equal to $$\frac{\pi}{4}+\pi n \text{, }n\in\mathbb{Z}$$but will $$\tan(x))^{\ln(\sin(x))}$$ ever equal $$(\ln(\cot(x)))^{\cos(x)}?$$

Here is my attempt at trying to figure it out:$$(\tan(x))^{\ln(\sin(x))}=(\ln(\cot(x)))^{\cos(x)}$$$$\ln(\sin(x))\ln(\tan(x))=\cos(x)\ln(\ln(\cot(x)$$$$\sec(x)\ln(\sin(x))\ln(\tan(x))=\ln(\ln(\cot(x)))$$So we need to see if the following equations will ever equal each other, and when I plug it into Desmos: Oh. I guess it doesn't intersect. But here's the thing: This might only be on the real numbers, so I can just check with Wolfram Alpha and get something, right? Right? There is no information given from Wolfram Alpha either, therefore, $(\tan(x))^{\ln(\sin(x))}\text{ will never equal }(\ln(\cot(x))^{\cos(x)}$

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My question.

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Is the conclusion that I have concluded with correct, or what could I do to attain the correct conclusion, or what could I do to attain it more quickly?

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Stuff that I might have messed up

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1. Taking the natural logarithms.
2. Simplifying the results.
3. Graphing it correctly.
4. The question tags (I sometimes mess that up)

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To clarify

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The whole "looking at previously asked questions and putting a twist to them" thing is not going to be an all the time thing, I am only doing this because there aren't any math problems that I either can't solve or have already solved at the moment.

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Imgur links in this post (Most likely not needed but still)

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Answer 1

One way to see the complex zeros is to plot$\arg(f(z))=\arg\left((\tan(z))^{\ln(\sin(z))}-(\ln(\cot(z))^{\cos(z)}\right),z=x+yi$:

The white line on the real imaginary axes support there are no pure imaginary roots. The zeros appear where many contour lines intersect. We notice there are infinite complex zeros symmetric across the imaginary axis and periodic by $2\pi$ with respect to the real axis due to $f(z)$’s trigonometric functions

Interestingly, there are a few more roots near $0$ but none are at $(x,0)$:

Analytically finding the roots would be hard from such a complicated equation.

Answer 2

This is an interesting equation to solve that is to say : find the zero of function $$f(x)=\big(\tan(x)\big)^{\log\left(\sin(x)\right)}-\big(\log(\cot(x))\big)^{\cos(x)}$$

I made $1,000$ random runs and, as you observed, I did not obtain a single solution (but this is not a proof).

Now, working in the complex domain : just copy/paste the next line for Wolfram Alpha

FindRoot[Tan[x]^Log[Sin[x]]==Log[Cot[x]]^Cos[x],{x,Pi+I E]

changing the starting value (here : $x_0=\pi + e\,i$)

Yous start from $\color{red}{\text{anywhere}}$ and you will arrive $\color{red}{\text{somewhere}}$.

Now, have fun.