Question

In a triangle $ABC,$ if $\frac{a^2+b^2}{a^2-b^2}\sin (A-B)=1$ and the triangle is not right angled, then $\cos (A-B)=$

• A. $\sin (\frac{C}{2}-\frac{\pi }{4})$
• B. $\cos (\frac{C}{2}-\frac{\pi }{4})$
• C. $\tan (\frac{C}{2}-\frac{\pi }{4})$
• D. $\cot (\frac{C}{2}-\frac{\pi }{4})$

Answer 1

The correct option is C $\tan (\frac{C}{2}-\frac{\pi }{4})$

$\Rightarrow \frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A-{\sin }^{2}B}\cdot \sin (A-B)=1$
$\Rightarrow \frac{{\sin }^{2}A+{\sin }^{2}B}{{\sin }^{2}A-{\sin }^{2}B}\times \frac{{\sin }^{2}A-{\sin }^{2}B}{\sin (A+B)}=1$
$(\because (\sin (A+B)\left)\right(\sin (A-B))={\sin }^{2}A-{\sin }^{2}B)$
$\Rightarrow \frac{{\sin }^{2}A+{\sin }^{2}B}{\sin (A+B)}=1$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=\sin (\pi -C)$
$\Rightarrow {\sin }^{2}A+{\sin }^{2}B=\sin C$
$\Rightarrow 1-{\cos }^{2}A+{\sin }^{2}B=\sin C$
$\Rightarrow 1-\sin C={\cos }^{2}A-{\sin }^{2}B$
$\Rightarrow \cos (A+B)\cos (A-B)=(1-\sin C)$
$\Rightarrow \cos (A-B)=\frac{1-\sin C}{\cos (\pi -C)}=\frac{1-\sin C}{-\cos C}$
$\therefore \cos (A-B)=\frac{1-\cos (\frac{\pi }{2}-C)}{-\sin (\frac{\pi }{2}-C)}=\frac{2{\sin }^2(\frac{\pi }{4}-\frac{C}{2})}{-2\sin (\frac{\pi }{4}-\frac{C}{2})\cos (\frac{\pi }{4}-\frac{C}{2})}=\tan (\frac{C}{2}-\frac{\pi }{4})$