Question

In $△ABC$ $\cos A+2\cos B+\cos C=2$ then

• A. $\tan \frac{A}{2}\tan \frac{C}{2}=3$
• B. $\cot \frac{A}{2}\cot \frac{C}{2}=3$
• C. $\cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$
• D. $\tan \frac{A}{2}\tan \frac{C}{2}=0$

Answer 1

Answer: (B), (C)

$\cot \frac{A}{2}\cot \frac{C}{2}=3$
$\cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$

$\cos A+2\cos B+\cos C=2$
$\Rightarrow \cos A+\cos C=2(1-\cos B)$
$\cos A+\cos C=4{\sin }^2\frac{B}{2}$
$\Rightarrow \cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)=2{\sin }^2\frac{B}{2}$
$\Rightarrow \cos \left(\frac{A+C}{2}\right).\cos \left(\frac{A-C}{2}\right)=2{\cos }^2\left(\frac{A+C}{2}\right)$
$\Rightarrow \frac{\cos \left(\frac{A-C}{2}\right)}{\cos \frac{A+C}{2}}=\frac{2}{1}$
Applying componendo and dividendo, we get
$\Rightarrow \frac{\cos \left(\frac{A-C}{2}\right)+\cos \left(\frac{A+C}{2}\right)}{\cos \left(\frac{A-C}{2}\right)-\cos \left(\frac{A-C}{2}\right)}=\frac{2+1}{2-1}$
$\Rightarrow \cot \frac{A}{2}.\cot \frac{C}{2}=\frac{3}{1}$ .....(1)
We know that
$\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
$\Rightarrow \cot \frac{A}{2}+\cot \frac{C}{2}=\cot \frac{B}{2}(\cot \frac{A}{2}.\cot \frac{C}{2}-1)$
$\Rightarrow \cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$ (by (1))