Question

If A,B,C be the angles of triangle ABC,such that A = $\frac{\pi }{4}$ and $\tan B\tan C=$ $\lambda$, then prove that
$\lambda <(\sqrt{2}-1{)}^{2}or\lambda >(\sqrt{2}+1{)}^2$

Answer 1

Since $A=\frac{\pi }{4}\therefore B+C=\frac{3\pi }{4}$

$\tan B\tan C=\lambda \Rightarrow \tan B\tan (\frac{3\pi }{4}-B)=\lambda$

or $\tan B\frac{(-1-\tan B)}{1-\tan B}=\lambda \therefore \tan \frac{3\pi }{4}=-1$

or ${\tan }^{2}B+(1-\lambda )\tan B+\lambda =0$

$\Delta \ge 0\Rightarrow (1-\lambda {)}^2-4\lambda \ge 0$

or ${\lambda }^2-6\lambda +1\ge 0$

Its roots are $a=3-2\sqrt{2},b=3+2\sqrt{2}$

$(\lambda -a)(\lambda -b)\ge 0,a<b$

$\therefore \lambda \le aor\lambda \ge b$

where $a=(\sqrt{2}-1{)}^2,b=(\sqrt{2}+1{)}^2$