Question

If in a triangle ABC, 5cosC + 6cosB = 4 and 6cosA + 4cosC = 5, then tan $\frac{A}{2}tan\frac{B}{2}$is equal to

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

Answer 1

Answer: (C)

Given : $5cosC+6cosB=4$ ...(1)
$6cosA+4cosC=5$ ...(2)
Adding eq. (1) & (2), we have
$9cosC+6(cosA+cosB)=9$
$\Rightarrow 9cosC+6[2cos\frac{A+B}{2},cos\frac{A-B}{2}]=9$
$\Rightarrow 9cosC-9+12cos(\frac{\pi }{2}-\frac{C}{2}).cos\frac{A-B}{2}=0$
$\Rightarrow 9(cosC-1)+12sin\frac{C}{2}.cos\frac{A-B}{2}=0$
$\Rightarrow 9[1-2si{n}^2\frac{C}{2}-1]+12sin\frac{C}{2}.cos\frac{A-B}{2}=0$
$\Rightarrow -18si{n}^2\frac{C}{2}+12sin\frac{C}{2}.cos\frac{A-B}{2}=0$
$\Rightarrow 3sin\frac{C}{2}=2cos\frac{A-B}{2}$
$\Rightarrow 3cos\frac{A+B}{2}=2cos\frac{A-B}{2}$
$\Rightarrow 3(cos\frac{A}{2}.cos\frac{B}{2}-sin\frac{A}{2}.sin\frac{B}{2})$
$\Rightarrow 2[cos\frac{A}{2}.cos\frac{B}{2}+sin\frac{A}{2}.sin\frac{B}{2}]$
$\Rightarrow 5sin\frac{A}{2}.sin\frac{B}{2}=cos\frac{A}{2}.cos\frac{B}{2}$
$\Rightarrow 5tan\frac{A}{2}.tancfracB2=1$
$\Rightarrow 5tan\frac{A}{2}.tan\frac{B}{2}=1$
$\Rightarrow tan\frac{A}{2}.tan\frac{B}{2}=\frac{1}{5}$