\[ \tan( a) + \cot( 2a) =0 \]

Solving the equation by converting (\(\cot 2a \)) to (\( \tan\)) gives No solution but when transposing cot 2a (and changing to tan) to other R.H.S yields two solutions?

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## Comments

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TopNewestCan you show your working so other others can see how you got 0 / 2 solutions?

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Sure.

\(tan\quad (x)\quad +\quad cot\quad (2x)\quad =\quad 0\\ tan\quad x\quad +\quad \frac { 1-{ \quad tan }^{ 2 }x\quad }{ 2tan\quad x } =\quad 0\\ { tan }^{ 2 }\quad x\quad +\quad 1\quad =\quad 0\\ { tan }^{ 2 }\quad x\quad =\quad -1\\ No\quad Solution\quad \quad \\ \\ tan\quad x\quad =\quad tan\quad (90\quad +\quad 2x)\\ \quad x\quad =\quad 2n\pi \quad \pm \quad 90\quad +\quad 2x\\ \\ \quad x\quad =\quad (\frac { 2n\pi }{ 3 } \quad -\quad 30)\\ or\\ \\ x\quad =\quad -(2n\pi \quad +\quad 90)\)

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First solution is correct i.e. no solution.

Your first step in the second solution is incorrect. Just because \(\tan(90+x)=-\cot(x)\) does not mean \(\tan(90+2x)=-\cot(2x)\).

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Can you list out what those values are?

What are the values of \( \tan x \)?

Hence, what is the conclusion? What mistake was made?

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\( \tan ( 90 ^ \circ + x ) = - \cot x \Leftrightarrow \tan (90 ^ \circ + 2 \theta ) = - \cot 2 \theta \).

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Edit : I agree that there is no solution but not with this solution.

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Yes, it is easier to prove it for the case when the value is actue. However, it is true for a wider set of values, namely \( \mathbb{R} \).

Look like you didn't read my other comment.

Basically, in the second solution, the values that you calculated are of the form \( 90 ^ \circ, 270^ \circ, 450^\circ , \ldots \). At these points, \( \tan x = \infty \), which is why they are not really considered solutions.

Similary, in your first approach, when you multiplied by \( \tan x \), you forgot to account for when \( \tan x = 0 , \infty \), as those could potentially introduce extraneous solutions to your equation.

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If you notice, in the second solution, tan(x) isn't defined.

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