If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$.

Question

If $\cot(x)=2$ and $\cot(y)=3$ show that $x+y=\frac{\pi}{4}$. I was trying to find x and y but is some weird thing. Also note that $x,y\:\in (0,\pi/2)$ Second idea was to write : $\frac{\cos x}{\sin x}=2$ and $\frac{\cos y}{\sin y}=3$ and adding this two I'll get $\frac{\cos(x+y)}{\cos x \cos y}=5$. Than what?

Answer 1

Hint: using basic trigonometric identitites:$$\tan(x+y)=\frac{\frac12+\frac13}{1-\frac12\frac13}=1\;\implies\;\ldots$$

Answer 2

HINT: We have $$\cot x=2\implies x=\tan^{-1}\left(\frac12\right)$$ and $$\cot y=3\implies y=\tan^{-1}\left(\frac13\right)$$ Now use the addition formula to show that $$\tan(x+y)=1$$

Answer 3

Alt. hint: $\;x,y$ are the angles marked in red below.

Answer 4

From this wikipedia page on trigonmetric identities, we see that

$\tan(\alpha + \beta) = \dfrac{(\tan \alpha + \tan \beta)}{1 - (\tan \alpha)(\tan \beta)}; \tag 1$

with

$\cot x = 2, \; \cot y = 3, \tag 2$

we find

$\tan x= \dfrac{1}{2} = 2^{-1}, \; \tan y = \dfrac{1}{3} = 3^{-1}; \tag 3$

thus

$\tan(x + y) = \dfrac{2^{-1} + 3^{-1}}{1 - 2^{-1} \cdot 3^{-1}} = \dfrac{5 \cdot 6^{-1}}{1 - 6^{-1}} = \dfrac{5 \cdot 6^{-1}}{5 \cdot 6^{-1}} = 1; \tag 4$

with $x, y \in (0, \pi / 2)$, we have $x + y \in (0, \pi)$; the only value of $x + y$ in this range satisfying (4) is

$x + y = \dfrac{\pi}{4}. \tag 5$