Question

In a $△ABC$ is $5\cos C+6\cos B=4$ and $6\cos A+4\cos C=5$, then $\tan \frac{A}{2}\tan \frac{B}{2}$=

• A. $\frac{1}{5}$
• B. $\frac{2}{5}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

Answer: (A)

$\frac{1}{5}$

In $△ABC,$

Given $5\cos C+6\cos B=4,6\cos A+4\cos C=5$

These are in the form of $b\cos C+c\cos B=a,c\cos A+a\cos C=b$

$⟹a=4,b=5,c=6$

$\tan \frac{A}{2}\tan \frac{B}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\times \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=\frac{s-c}{s}=\frac{2s-2c}{2s}=\frac{a+b-c}{a+b+c}=\frac{4+5-6}{4+5+6}=\frac{3}{15}=\frac{1}{5}$