Question

If in $△ABC,\angle A=\frac{\pi }{4}$ and $\tan B\tan C=p$, then the possible set of value(s) of $p$ is/are

• A. $(-\infty ,0)$
• B. $\left(\right(\sqrt{2}+1{)}^2,\infty )$
• C. $\left[\right(\sqrt{2}+1{)}^2,\infty )$
• D. $(-\infty ,0]$

Answer 1

Answer: (A), (C)

The correct options are
A $(-\infty ,0)$
C $\left[\right(\sqrt{2}+1{)}^2,\infty )$

Given: $A+B+C=\pi ,\angle A=\frac{\pi }{4}$
$\Rightarrow B+C=\frac{3\pi }{4}\Rightarrow \frac{3\pi }{4}>B,C>0$
Also,
$\tan B\tan C=p\Rightarrow \frac{\sin B\sin C}{\cos B\cos C}=\frac{p}{1}\Rightarrow \frac{\cos B\cos C-\sin B\sin C}{\cos B\cos C+\sin B\sin C}=\frac{1-p}{1+p}\Rightarrow \frac{\cos (B+C)}{\cos (B-C)}=\frac{1+p}{1-p}\Rightarrow -\frac{1}{\sqrt{2}\cos (B-C)}=\frac{1+p}{1-p}\Rightarrow \cos (B-C)=\frac{1+p}{\sqrt{2}(p-1)}$

Now, we know that
$\frac{3\pi }{4}>B-C\ge 0\Rightarrow 1\ge \cos (B-C)>-\frac{1}{\sqrt{2}}\Rightarrow 1\ge \frac{1+p}{\sqrt{2}(p-1)}>-\frac{1}{\sqrt{2}}\Rightarrow \sqrt{2}\ge \frac{1+p}{p-1}>-1$
Solving both inequality, we get
$\frac{1+p}{p-1}>-1\Rightarrow \frac{1+p}{p-1}+1>0\Rightarrow \frac{2p}{p-1}>0\Rightarrow p\in (-\infty ,0)\cup (1,\infty )\cdots \left(1\right)$

$\sqrt{2}\ge \frac{1+p}{p-1}\Rightarrow \frac{1+p}{p-1}-\sqrt{2}\le 0\Rightarrow \frac{p(1-\sqrt{2})+(1+\sqrt{2})}{p-1}\le 0\Rightarrow \frac{-p+(\sqrt{2}+1{)}^2}{p-1}\le 0\Rightarrow \frac{p-(\sqrt{2}+1{)}^2}{p-1}\ge 0\Rightarrow p\in (-\infty ,0)\cup \left[\right(\sqrt{2}+1{)}^2,\infty )\cdots (2)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$p\in (-\infty ,0)\cup \left[\right(\sqrt{2}+1{)}^2,\infty )$