# Two solutions or No solution?

$\tan( a) + \cot( 2a) =0$

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?

Please help! Note by Vishal Yadav
5 years, 2 months ago

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## Comments

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

Staff - 5 years, 2 months ago

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Sure.

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

- 5 years, 2 months ago

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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 $\tan(90+2x)=-\cot(2x)$.

- 5 years, 2 months ago

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The first one is mine and the second one is the Author's (RD Sharma's) solution. I didn't expect his solution to be incorrect( as he is an renowned mathematician) so I had this note.

- 5 years, 2 months ago

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Consider the 2nd solution.

Can you list out what those values are?

What are the values of $\tan x$?

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

Staff - 5 years, 2 months ago

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I disagree with your claim in the second line.

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

Staff - 5 years, 2 months ago

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Since we don't know the range of $x$ so we cannot state whether 2x is acute. If 2x is obtuse (90 +2x) lies on third quadrant and Cot 2x is $+$.

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

- 5 years, 2 months ago

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Nope. $x$ being acute has nothing to do with this. This statement is about translation and reflection of the tangent graph to obtain the cotangent graph. Alternatively, you could use unit circle trigonometry to prove the statement.

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.

Staff - 5 years, 2 months ago

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I feel Iike I need to know trigonometry a bit more. :P

- 5 years, 2 months ago

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

- 5 years, 2 months ago

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