Why Does This Logarithmic Trigonometric Equation Equal Zero?

[Manasan3010]

Homework Statement

Prove,
$\frac{1}{\log_{tan\ 1^\circ}(2021)} +\frac{1}{\log_{tan\ 2^\circ}(2021)} +\frac{1}{\log_{tan\ 3^\circ}(2021)} +...+\frac{1}{\log_{tan\ 89^\circ}(2021)} = 0$

Relevant Equations

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The attempt at a solution

${\log_{2021}(tan\ 1^\circ)} +{\log_{2021}(tan\ 2^\circ)} +{\log_{2021}(tan\ 3^\circ)} +...+{\log_{2021}(tan\ 89^\circ)} = 0 \\ Antilog_{2021}[{\log_{2021}(tan\ 1^\circ)} +{\log_{2021}(tan\ 2^\circ)} +{\log_{2021}(tan\ 3^\circ)} +...+{\log_{2021}(tan\ 89^\circ)} ] = Antilog_{2021}[0]\\ tan\ 1^\circ + tan\ 2^\circ + tan\ 3^\circ +...+ tan\ 89^\circ = 1\\ arctan(tan\ 1^\circ + tan\ 2^\circ + tan\ 3^\circ +...+ tan\ 89^\circ = 1)\\ 1^\circ +2^\circ +3^\circ +...+89^\circ = 45^\circ \\ \frac{89}{2} \times 90^\circ=45^\circ \\$
Where did I make mistake?
I also want to know How to display a long equation in the same line in MathJax(Ex. 2nd line)

 

[Mark44]

Problem Statement: Proof,

The above should be Prove.
You prove (verb) a statement to arrive at a proof (noun).

$\frac{1}{\log_{tan\ 1^\circ}(2021)} +\frac{1}{\log_{tan\ 2^\circ}(2021)} +\frac{1}{\log_{tan\ 3^\circ}(2021)} +...+\frac{1}{\log_{tan\ 89^\circ}(2021)} = 0$
Relevant Equations: -

Working

${\log_{2021}(tan\ 1^\circ)} +{\log_{2021}(tan\ 2^\circ)} +{\log_{2021}(tan\ 3^\circ)} +...+{\log_{2021}(tan\ 89^\circ)} = 0 \\ Antilog_{2021}[{\log_{2021}(tan\ 1^\circ)} +{\log_{2021}(tan\ 2^\circ)} +{\log_{2021}(tan\ 3^\circ)} +...+{\log_{2021}(tan\ 89^\circ)} ] = Antilog_{2021}[0]$

The line below has an error.
$f(a + b + c) \ne f(a) + f(b) + f(c)$, in general.

$tan\ 1^\circ + tan\ 2^\circ + tan\ 3^\circ +...+ tan\ 89^\circ = 1\\ arctan(tan\ 1^\circ + tan\ 2^\circ + tan\ 3^\circ +...+ tan\ 89^\circ = 1)\\ 1^\circ +2^\circ +3^\circ +...+89^\circ = 45^\circ \\ \frac{89}{2} \times 90^\circ=45^\circ \\$
Where did I make mistake?
I also want to know How to display a long equation in the same line in MathJax(Ex. 2nd line)

 

[Manasan3010]

The line below has an error.
f(a+b+c)≠f(a)+f(b)+f(c)f(a+b+c)≠f(a)+f(b)+f(c)f(a + b + c) \ne f(a) + f(b) + f(c), in general.

Which line are you referring to and What should I to get the correct steps?

 

[Mark44]

Which line are you referring to and What should I to get the correct steps?

The second and third lines of your work.

$Antilog_{2021}[{\log_{2021}(tan\ 1^\circ)} +{\log_{2021}(tan\ 2^\circ)} +{\log_{2021}(tan\ 3^\circ)} +...+{\log_{2021}(tan\ 89^\circ)} ] = Antilog_{2021}[0]$

The line below doesn't follow from the line above.

$tan\ 1^\circ + tan\ 2^\circ + tan\ 3^\circ +...+ tan\ 89^\circ = 1$

You are in essence saying that $f(a + b + c) = f(a) + f(b) + f(c)$, which is generally not true.

Start in again with your first line:
$\log_{2021}(\tan 1^\circ) +\log_{2021}(\tan 2^\circ) +\log_{2021}(\tan 3^\circ) +...+\log_{2021}(\tan 89^\circ) = 0$

Use the idea that $\log_b(A) + \log_b(B) = \log_b(AB)$

 

[Manasan3010]

I found the solution Thanks.

If anyone looking for solution Then,
tan1 .tan2 .tan3……tan87 .tan88. tan89

tan1 . tan 2 . tan 3 ….. cot3. cot 2. cot 1

tan1.cot1 . tan2.cot2 …..tan44.cot44. tan45

1*1*1*1…….1

 

[vela]

There's actually an issue with the problem as stated. The middle term of the series would be
$$\frac 1{\log_{\tan 45^\circ}2021},$$ but the logarithm is undefined since there's no power of 1 that's equal to 2021.