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  4. If in a triangle ABC, 5cosC + 6cosB = 4 and 6cosA + 4cosC = 5, then tan (A/2) tan(B/2)is equal to

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

VITEEEVITEEE 2006

Solution:

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