Question

If $a=5,b=4$ and $\cos (A-B)=\frac{31}{32},$ then $\tan \frac{C}{2}=$

• A. $\frac{\sqrt{7}}{3}$
• B. $\frac{3}{\sqrt{7}}$
• C. $\frac{1}{\sqrt{7}}$
• D. $\frac{2}{\sqrt{7}}$

Answer 1

The correct option is A $\frac{\sqrt{7}}{3}$

According to given ratios, let $\sin A=5\lambda ,\sin B=4\lambda$
Given: $\cos (A-B)=\frac{31}{32}(\text{Let}\tan \frac{(A-B)}{2}=t)$
$\Rightarrow \frac{1-t^2}{1+t^2}=\frac{31}{32}\Rightarrow t=\frac{1}{3\sqrt{7}}$
We know, $\left(\frac{a-b}{a+b}\right)\cot \frac{C}{2}=\tan \frac{(A-B)}{2}$
$\Rightarrow \frac{1}{9}\cot \frac{C}{2}=\frac{1}{3\sqrt{7}}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$