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

Question

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?

https://math.stackexchange.com/questions/1861288/on-the-proof-tan-70-tan-20-2-tan-40-4-tan-10 — lab bhattacharjee, Apr 15 '18 at 17:39

score 2 · Answer 1 · answered Apr 15 '18 at 17:56

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

answered Apr 15 '18 at 17:56

kayush

2,441

We need to deal with $$\tan40^\circ+4\tan10^\circ$$ and not $$\tan40^\circ+2\tan10^\circ$$ right? – lab bhattacharjee Apr 16 '18 at 06:13

score 0 · Answer 2 · answered Apr 15 '18 at 17:43

0

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

answered Apr 15 '18 at 17:43

Dr. Sonnhard Graubner

95,283

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

2 Answers2