$\tan (40^{\circ})+4\tan (10^{\circ})=\cot(x^{\circ})$ Find $x$

Question

$\tan (40^{\circ})+4\tan (10^{\circ})=\cot(x^{\circ})$ Find $x$

My Try:

we have $\tan x-\cot x=-2 \cot 2x$

Subtracted both sides with $\cot (40)$ we get

$$\tan 40-\cot 40+4 \tan 10=\cot x-\cot 40$$ that is

$$-2 \cot 80+4 \tan 10=\cot x-\cot 40$$ $\implies$

$$2 \tan 10+\cot 40=\cot x$$

any clue here?

Answer 1

Hint: use that $$\tan(4x)=\tan(2x+2x)=\frac{2\tan(2x)}{1-\tan^2(2x)}$$

Answer 2

Just few steps needed after what you have done to complete

$$\tan 10 + \tan 10 + \tan 40 = \cot x$$ $$\tan 10 + \tan 50 (1- \tan 40 \tan 10) = \cot x$$ $$\tan 10 + \tan 50 - (\tan50 \tan 40) \tan 10 = \cot x$$ $$\tan 10 + \tan 50 - \tan 10 = \cot x$$ $$\tan 50 = \cot 40 = \cot x$$

if $0\leq x \leq 90 \implies x = 40$