Question

Let A,B,C be three angles such that $A=\frac{\pi }{4}$ and $\tan B\tan C=p$. Find all possible values of p such that A,B,C are the angles of a triangle.

• A. $p<0$ or $p>{(\sqrt{2}+1)}^2$
• B. $p>0$ or $p<{(\sqrt{2}+1)}^2$
• C. $p>0$ or $p<{(\sqrt{2}-1)}^2$
• D. $p<0$ or $p>{(\sqrt{2}-1)}^2$

Answer 1

Answer: (A)

$p<0$ or $p>{(\sqrt{2}+1)}^2$

$A+B+C=\pi \Rightarrow B+C=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$ ...(1)

$(\because A=\frac{\pi }{4})$

$\tan B.\tan C=p\Rightarrow \frac{\sin B.\sin C}{\cos B.\cos C}=\frac{p}{1}$

$\Rightarrow \frac{\sin B.\sin C+\cos B.\cos C}{\sin B.\sin C-\cos B.\cos C}=\frac{p+1}{p-1}$

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

$\Rightarrow \cos (B-C)=\frac{(1+p)}{\sqrt{2}(1-p)}$ ...(2)

$(\because B+C=\frac{3\pi }{4})$

Since $B$ or $C$ can vary from $0$ to $\frac{3\pi }{4}$

$0\le B-C<\frac{3\pi }{4}\Rightarrow -\frac{1}{\sqrt{2}}<\cos (B-C)\le 1$ ...(3)

From (2) and (3), we get

$-\frac{1}{\sqrt{2}}<\frac{1+p}{\sqrt{2}(p-1)}\le 1$

$\Rightarrow -\frac{1}{\sqrt{2}}<\frac{1+p}{\sqrt{2}(p-1)}$ and $\frac{1+p}{\sqrt{2}(p-1)}\le 1$

$\Rightarrow \frac{1+p}{p-1}+1\ge 0$ and $\frac{1+p-\sqrt{2}p+\sqrt{2}}{\sqrt{2}(p-1)}\le 0$

$\Rightarrow \frac{2p}{p-1}\ge 0$ and $\frac{(1-\sqrt{2})(p-\frac{1+\sqrt{2}}{1-\sqrt{2}})}{\sqrt{2}(p-1)}\le 0$

$\Rightarrow \frac{2p}{p-1}>0$ and $\frac{\left({p-(\sqrt{2}+1)}^2\right)}{(p-1)}\ge 0$

$\Rightarrow (p<0orp>1)$ and $(p<1orp>{(\sqrt{2}+1)}^2)$

$p<0$ or $p\ge {(\sqrt{2}+1)}^2$