I have to solve the following equation
$$5\tan(2\theta)=4\cot(\theta),\space \space \space 0°<\theta<180°$$
Applying the double angle formula for $\tan(2\theta)$
$$\frac{10\tan(\theta)}{1-\tan^2(\theta)}=4\cot(\theta)$$
I did solve and simplify this further but did not arrive at a correct value for $\theta$ so I decided to confirm where I went wrong on wolfram alpha, and to my surprise my very first step turned out to be wrong.
What mistake have I made here, is the double angle formula I have applied, $\tan(2\theta)=\dfrac{2\tan(\theta)}{1-\tan^2(\theta)}$, wrong or is there some restriction upon applying it?
Best Answer
HINT
The identity which you have applied is correct. More precisely, we have that \begin{align*} \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{2\sin(\theta)\cos(\theta)}{\cos^{2}(\theta) - \sin^{2}(\theta)} = \frac{2\tan(\theta)}{1 - \tan^{2}(\theta)} \end{align*}
Consequently, the proposed equation is equivalent to \begin{align*} 5\tan(2\theta) = 4\cot(\theta) \Longleftrightarrow \frac{10\tan(\theta)}{1 - \tan^{2}(\theta)} = \frac{4}{\tan(\theta)} \end{align*}
Can you take it from here?