In-ABC-if-cot-A-cot-C-1-2-cot-B-cot-C-1-18-then-tan-C-

Question Number 164426 by cortano1 last updated on 17/Jan/22

I n Δ A B C i f {\begin{cases} \cot A . \cot C = \frac{1}{2} \\ \cot B . \cot C = \frac{1}{18} \end{cases}

t h e n \tan C = ?

Commented by bobhans last updated on 17/Jan/22

let \tan C = x > 0

\Rightarrow \frac{2}{x} + \frac{18}{x} + x = \frac{2}{x} . \frac{18}{x} . x

\Rightarrow 20 + x^{2} = 36 \Rightarrow x = \tan C = 4 ✓

Answered by som(math1967) last updated on 17/Jan/22

A + B + C = π

C = π - (A + B)

c o t C = - c o t (A + B)

c o t C = - \frac{c o t A c o t B - 1}{c o t B + c o t A}

c o t B c o t C + c o t A c o t C = - c o t A c o t B + 11

\Rightarrow c o t A c o t B = 1 - \frac{1}{18} - \frac{1}{2} = \frac{18 - 1 - 9}{18} = \frac{8}{18} = \frac{4}{9}

{c o t}^{2} {A c o t}^{2} {B c o t}^{2} C = \frac{1}{2} \times \frac{1}{18} \times \frac{4}{9} = \frac{1}{81}

c o t A c o t B c o t C = \frac{1}{9}

t a n C = \frac{1}{c o t C} = \frac{c o t A c o t B}{c o t A c o t B c o t C} = \frac{4}{9} \times \frac{9}{1}

t a n C = 4

Commented by cortano1 last updated on 17/Jan/22

o n l y 4 s i r

Commented by som(math1967) last updated on 17/Jan/22

y e s t a n C > 0