Trigonometry #134

$For\quad x+y+z=\pi ,\quad we\quad know\quad that\\ tan\quad x\quad +\quad tan\quad y\quad +\quad tan\quad z\quad =\quad tan\quad x\quad \times \quad tan\quad y\quad \times \quad tan\quad z\\ \\ We\quad know\quad tan\quad x\quad \times \quad tan\quad z\quad =\quad 2\\ \therefore \quad tan\quad x\quad +\quad tan\quad y\quad +\quad tan\quad z\quad =\quad 2tan\quad y\\ =>\quad tan\quad x\quad +\quad tan\quad z\quad =\quad tan\quad y\quad \quad ->\quad eq(1)\\ \\ We\quad know\quad tan\quad y\quad \times \quad tan\quad z\quad =\quad 18\\ \quad \therefore \quad tan\quad x\quad +\quad tan\quad y\quad +\quad tan\quad z\quad =\quad 18tan\quad x\\ =>\quad tan\quad y\quad +\quad tanz\quad =\quad 17tan\quad x\quad ->eq(2)\\ From\quad eq\quad (1)\quad and\quad eq(2),\quad we\quad get\quad tan\quad z\quad =\quad 8tan\quad x\\ \\ \therefore \quad tan\quad x\quad =\quad \frac { 1 }{ 2 } \quad =>\quad tan\quad z\quad =\quad 4\quad \quad \\ \boxed { \therefore \quad { tan }^{ 2 }z\quad =\quad 16\quad }$

For $x+y+z=\pi$ , we know that $\cot x \cot y + \cot y \cot z + \cot x \cot z = 1$ Given $\tan x \tan z = 2$ and $\tan y \tan z = 18$ , we get $\cot x \cot z = \frac{1}{2}$ and $\cot y \cot z = \frac{1}{18}$ . Hence,

$\cot x \cot y + \frac{1}{18} + \frac{1}{2} = 1 \\ \cot x \cot y + \frac{5}{9} = 1 \\ \cot x \cot y = \frac{4}{9}$

Now, we have the following equations. $\cot x \cot y = \frac{4}{9} \\ \cot y \cot z = \frac{1}{18} \\ \cot z \cot x = \frac{1}{2}$ We shall solve these simultaneously to find the value of $\cot z$ .

$\cot x = \frac{1}{2\cot z}~ and ~\cot y = \frac{1}{18\cot z} \\ \therefore \left(\frac{1}{2\cot z}\right)\left(\frac{1}{18\cot z}\right)=\frac{4}{9} \\ \frac{1}{36\cot^{2}z}=\frac{4}{9} \\ \frac{1}{\cot^{2}z}=16 \\ \boxed{\tan^{2}z=16}$

Given that $\begin{cases} \tan x \tan z = 2 & ...(1) \\ \tan y \tan z = 18 & ...(2) \end{cases}$

$\dfrac {(2)}{(1)}: \quad \dfrac {\tan y \tan z}{\tan x \tan z} = \dfrac {\tan y}{\tan x} = \dfrac {18}2 = 9 \implies \color{#D61F06}\tan y = 9 \tan x$

Since $x+y+z=\pi$ :

$\begin{aligned} \implies \tan x {\color{#3D99F6}\tan y \tan z} & = \tan x + {\color{#D61F06}\tan y} + \tan z & \small \color{#3D99F6} \text{Given that }\tan y \tan z = 18 \\ {\color{#3D99F6}18} \tan x & = \tan x + {\color{#D61F06}9\tan x} + \tan z & \small \color{#D61F06} \text{Note that }\tan y = 9\tan x \\ \implies \tan z & = 8 \tan x \end{aligned}$

Now consider,

$\begin{aligned} \tan \color{#3D99F6} z & = \tan ({\color{#3D99F6} \pi - x - y}) & \small \color{#3D99F6} \text{Since }x+y+z=\pi \\ & = - \tan (x+y) & \small \color{#3D99F6} \text{and } \tan (\pi - \theta) = - \tan \theta \\ \implies \color{#3D99F6} \tan z & = - \frac {\tan x + \tan y}{1-\tan x \tan y} & \small \color{#3D99F6} \tan z = 8 \tan x \\ \color{#3D99F6} 8 \tan x & = \frac {\tan x + 9\tan x}{9\tan^2 x - 1} \\ 4 & = \frac 5{9\tan^2 x - 1} \\ 36\tan^2 x - 4 & = 5 \\ \tan^2 x & = \frac 14 \\ \implies \tan^2 z & = (8 \tan x)^2 \\ & = 64 \times \frac 14 = \boxed {16} \end{aligned}$