Q56 / 120 · VITEEE 2006 · Mathematics
If in a triangle $\mathrm{ABC}$, $5 \cos C+6 \cos B=4$ and $6 \cos A+4 \cos C=5$ then $\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}$ is equal to
- A$\frac{2}{3}$
- B$\frac{3}{2}$
- Correct answerC$\frac{1}{5}$
- D5
Solution
Given : $5 \cos C+6 \cos B=4$ ...(1)
$6 \cos A+4 \cos C=5$ ...(2)
Adding eq. (1) \& (2), we have
$9 \cos C+6(\cos A+\cos B)=9$
$\Rightarrow 9 \cos \mathrm{C}+6\left[2 \cos \frac{\mathrm{A}+\mathrm{B}}{2} \cdot \cos \frac{\mathrm{A}-\mathrm{B}}{2}\right]=9$
$\Rightarrow 9 \cos \mathrm{C}-9+12 \cos \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right) \cdot \cos \frac{\mathrm{A}-\mathrm{B}}{2}=0$
$\Rightarrow 9(\cos \mathrm{C}-1)+12 \sin \frac{\mathrm{C}}{2} \cdot \cos \frac{\mathrm{A}-\mathrm{B}}{2}=0$
$\Rightarrow 9\left[1-2 \sin ^{2} \frac{\mathrm{C}}{2}-1\right]+12 \sin \frac{\mathrm{C}}{2} \cdot \cos \frac{\mathrm{A}-\mathrm{B}}{2}=0$
$\Rightarrow-18 \sin ^{2} \frac{\mathrm{C}}{2}+12 \sin \frac{\mathrm{C}}{2} \cdot \cos \frac{\mathrm{A}-\mathrm{B}}{2}=0$
$\Rightarrow 3 \sin \frac{\mathrm{C}}{2}=2 \cos \frac{\mathrm{A}-\mathrm{B}}{2}$
$\Rightarrow 3 \cos \frac{\mathrm{A}+\mathrm{B}}{2}=2 \cos \frac{\mathrm{A}-\mathrm{B}}{2}$
$\Rightarrow 3\left(\cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}-\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{B}}{2}\right)$
$=2\left[\cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}+\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{B}}{2}\right]$
$\Rightarrow 5 \sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{B}}{2}=\cos \frac{\mathrm{A}}{2} \cdot \cos \frac{\mathrm{B}}{2}$
$\Rightarrow 5 \tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2}=1$
$\Rightarrow \tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2}=\frac{1}{5}$
Hence, the correct answer is option: 'C'.