Question

If in a $△ABC,\cos A+2\cos B+\cos C=2$ then $a,b,c$ are in

• A. A.P
• B. G.P
• C. H.P
• D. none of these

Answer 1

The correct option is A A.P

Given:$\cos A+2\cos B+\cos C=2$
$\Rightarrow \cos A+\cos C=2-2\cos B$
Using transformation angle formulae, we have
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2(1-\cos B)$
Using multiple angle formula, we get
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2.2{\sin }^2\left(\frac{B}{2}\right)$
We have $A+B+C=\pi \Rightarrow .A+C=\pi -B\Rightarrow \frac{A+C}{2}=\frac{\pi }{2}-\frac{B}{2}$
$\Rightarrow 2\cos (\frac{\pi }{2}-\frac{B}{2})\cos \left(\frac{A-C}{2}\right)=2.2{\sin }^2\left(\frac{B}{2}\right)$
$\Rightarrow 2\sin \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)=2.2{\sin }^2\left(\frac{B}{2}\right)$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin \left(\frac{B}{2}\right)$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin (\frac{\pi }{2}-\frac{A+C}{2})$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\cos \left(\frac{A+C}{2}\right)$
$\Rightarrow \frac{\cos \left(\frac{A-C}{2}\right)}{\cos \left(\frac{A+C}{2}\right)}=\frac{2}{1}$
By componendo and dividendo method,
$\Rightarrow \frac{\cos \left(\frac{A-C}{2}\right)-\cos \left(\frac{A+C}{2}\right)}{\cos \left(\frac{A-C}{2}\right)+\cos \left(\frac{A+C}{2}\right)}=\frac{2-1}{2+1}$
$\Rightarrow \frac{-2\sin \left(\frac{A-C+A+C}{2}\right)\sin \left(\frac{A-C-A-C}{2}\right)}{2\cos \left(\frac{A-C+A+C}{2}\right)\cos \left(\frac{A-C-A-C}{2}\right)}=\frac{1}{3}$
$\Rightarrow \frac{\sin \left(\frac{A}{2}\right)\sin \left(\frac{C}{2}\right)}{\cos \left(\frac{A}{2}\right)\cos \left(\frac{C}{2}\right)}=\frac{1}{3}$
$\Rightarrow \tan \left(\frac{A}{2}\right)\tan \left(\frac{C}{2}\right)=\frac{1}{3}$
$\Rightarrow \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=\frac{1}{3}$
$\Rightarrow \frac{s-b}{s}=\frac{1}{3}$
$\Rightarrow 3s-3b=s$
$\Rightarrow 2s=3b$
We know that $a+b+c=2s$
$\therefore a+b+c=3b$
Hence,$a+c=2b$
$\therefore a,b,c$ are in A.P