Question

In a $△ABC,$ if $\left(\frac{a^2+b^2}{a^2-b^2}\right)\sin (A-B)=1$, then either $C$ is a right angle or $\cos (A-B)=\tan (\frac{C}{2}-\frac{\pi }{4})$.

• A. True
• B. False

Answer 1

The correct option is A True

$\left(\frac{a^2+b^2}{a^2-b^2}\right)\sin (A-B)=1$
Using sine rule
$\Rightarrow \frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A-{\sin }^{2}B}.\sin (A-B)=1$
$\Rightarrow \frac{{\sin }^{2}A+{\sin }^{2}B}{\sin (A+B)\sin (A-B)}.\sin (A-B)=1$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=\sin (A+B)$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=\sin (\pi -C)$
$\Rightarrow \frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}=\sin C$
$\Rightarrow 1-\cos 2A+1-\cos 2B=2\sin C$
$\Rightarrow \cos 2A+\cos 2B=2-2\sin C$
$\Rightarrow \cos 2A+\cos 2B=2(1-\sin C)$
$\Rightarrow 2\cos (A+B)\cos (A-B)=2(1-\sin C)$
$\Rightarrow \cos (\pi -C)\cos (A-B)=2(1-\sin C)$
$\Rightarrow -\cos C\cos (A-B)=2(1-\sin C)$
$\Rightarrow \cos (A-B)=\frac{2(1-\sin C)}{-\cos C}$
$\Rightarrow \cos (A-B)=\frac{2({\sin }^2\frac{C}{2}+{\cos }^2\frac{C}{2}-2\sin \frac{C}{2}\cos \frac{C}{2})}{-({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2})}$
$=\frac{{(\sin \frac{C}{2}-\cos \frac{C}{2})}^2}{({\sin }^2\frac{C}{2}-{\cos }^2\frac{C}{2})}$
$=\frac{{(\sin \frac{C}{2}-\cos \frac{C}{2})}^2}{(\sin \frac{C}{2}+\cos \frac{C}{2})(\sin \frac{C}{2}-\cos \frac{C}{2})}$
$=\frac{(\sin \frac{C}{2}-\cos \frac{C}{2})}{(\sin \frac{C}{2}+\cos \frac{C}{2})}$
Divide both sides by $\cos \frac{C}{2}$ we get
$=\frac{\tan \frac{C}{2}-1}{\tan \frac{C}{2}+1}$
Take $1=\tan \frac{\pi }{4}$
$=\frac{\tan \frac{C}{2}-\tan \frac{\pi }{4}}{\tan \frac{\pi }{4}\tan \frac{C}{2}+1}$
$=\tan (\frac{C}{2}-\frac{\pi }{4})$