Question

In a $△ABC$, if $\angle A=\frac{\pi }{4}$ and $\tan B\tan C=p$, then

• A. $p^2-6p+1\le 0$
• B. $p^2+6p-1\ge 0$
• C. $p^2-6p+1\ge 0$
• D. none of these

Answer 1

Answer: (C)

$p^2-6p+1\ge 0$

We know that, in $△ABC$

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\Rightarrow \tan B+\tan C=\tan A(\tan B\tan C-1)=\tan \frac{\pi }{4}(p-1)=(p-1)$

$\Rightarrow \tan B+\frac{p}{\tan B}=p-1\Rightarrow {\tan }^{2}B-(p-1)\tan B+p=0$

Since $\tan B$ is real, therefore

$\therefore {(p-1)}^2-4p\ge 0$

$\Rightarrow p^2-6p+1\ge 0$