Question

tg18=a. tg81=?
Please , can you helo me ?

Answer 1

Ответ:

$tg81=ctg9=\frac{\sqrt{1+a^2}+1}{a}$

Объяснение:

$tg81=tg(90-9)=ctg9=\frac{1}{tg9}\\tg18=tg(2*9)=\frac{2tg9}{1-tg^29}=a\\2tg9=a(1-tg^29)\\2tg9=a-atg^29\\atg^29+2tg9-a=0$

Solve quadratic equation for tg9:

$tg9=\frac{-2+-\sqrt{2^2-4*a*(-a)}}{2a}=\frac{-2+-\sqrt{4+4a^2}}{2a}=\frac{-2+-2\sqrt{1+a^2}}{2a}=\frac{-1+-\sqrt{1+a^2}}{a}$

As tg9 > 0:

$tg9=\frac{-1+\sqrt{1+a^2}}{a}$

$ctg9=\frac{1}{tg9}=\frac{a}{\sqrt{1+a^2}-1}=\frac{a(\sqrt{1+a^2}+1)}{1+a^2-1}=\frac{a(\sqrt{1+a^2}+1)}{a^2}=\frac{\sqrt{1+a^2}+1}{a}$