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

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!

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

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!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestCan you show your working so other others can see how you got 0 / 2 solutions? – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply

\(tan\quad (x)\quad +\quad cot\quad (2x)\quad =\quad 0\\ tan\quad x\quad +\quad \frac { 1-{ \quad tan }^{ 2 }x\quad }{ 2tan\quad x } =\quad 0\\ { tan }^{ 2 }\quad x\quad +\quad 1\quad =\quad 0\\ { tan }^{ 2 }\quad x\quad =\quad -1\\ No\quad Solution\quad \quad \\ \\ tan\quad x\quad =\quad tan\quad (90\quad +\quad 2x)\\ \quad x\quad =\quad 2n\pi \quad \pm \quad 90\quad +\quad 2x\\ \\ \quad x\quad =\quad (\frac { 2n\pi }{ 3 } \quad -\quad 30)\\ or\\ \\ x\quad =\quad -(2n\pi \quad +\quad 90)\) – Vishal Yadav · 1 year, 5 months ago

Log in to reply

– Deeparaj Bhat · 1 year, 5 months ago

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

Your first step in the second solution is incorrect. Just because \(\tan(90+x)=-\cot(x)\) does not mean \(\tan(90+2x)=-\cot(2x)\). – Brilliant Member · 1 year, 5 months ago

Log in to reply

\( \tan ( 90 ^ \circ + x ) = - \cot x \Leftrightarrow \tan (90 ^ \circ + 2 \theta ) = - \cot 2 \theta \). – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply

Edit : I agree that there is no solution but not with this solution. – Vishal Yadav · 1 year, 5 months ago

Log in to reply

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. – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply

– Vishal Yadav · 1 year, 5 months ago

I feel Iike I need to know trigonometry a bit more. :PLog in to reply

– Vishal Yadav · 1 year, 5 months ago

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.Log in to reply

Can you list out what those values are?

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

Hence, what is the conclusion? What mistake was made? – Calvin Lin Staff · 1 year, 5 months ago

Log in to reply