Question

Assertion :If in a triangle ,$\cos A+2\cos B+\cos C=2,$ then $a,b,c$ must be in A.P Reason: $\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

• A. Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion.
• B. Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion

$\cos A+2\cos B+\cos C=2$
$\Rightarrow \cos A+\cos C=2-2\cos B$
$\Rightarrow \cos A+\cos C=2(1-\cos B)$
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2(1-\cos B)$ using transformation angle formula
$\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2.2{\sin }^2\left(\frac{B}{2}\right)$ using multiple angle formula
$\Rightarrow 2\cos (\frac{\pi }{2}-\frac{B}{2})\cos \left(\frac{A-C}{2}\right)=2.2{\sin }^2\left(\frac{B}{2}\right)$ since $A+B+C=\pi$
$\Rightarrow 2\sin \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)=4{\sin }^2\left(\frac{B}{2}\right)$ since $\cos (\frac{\pi }{2}-\theta )=\sin \theta$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin \left(\frac{B}{2}\right)$ by cancelling the like terms
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin (\frac{\pi }{2}-\frac{A+C}{2})$ since $A+B+C=\pi$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\cos \left(\frac{A+C}{2}\right)$ since $\sin (\frac{\pi }{2}-\theta )=\cos \theta$
$\Rightarrow \frac{\cos \left(\frac{A-C}{2}\right)}{\cos \left(\frac{A+C}{2}\right)}=\frac{2}{1}$ by dividing both sides by $\cos \left(\frac{A+C}{2}\right)$
$\Rightarrow \frac{\cos \left(\frac{A+C}{2}\right)+\cos \left(\frac{A-C}{2}\right)}{\cos \left(\frac{A-C}{2}\right)-\cos \left(\frac{A+C}{2}\right)}=\frac{2+1}{2-1}$ using componendo dividendo rule.
$\therefore \frac{2\cos \frac{A}{2}\cos \frac{C}{2}}{2\sin \frac{A}{2}\sin \frac{C}{2}}=3$ using transformation angle formula
or $\cot \frac{A}{2}\cot \frac{C}{2}=3$ using the basic trigonometric identity $\cot \theta =\frac{\cos \theta }{\sin \theta }$
or $\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=3$
$\Rightarrow \frac{s}{s-b}=3$ by evaluating
$\Rightarrow s=3s-3b$
or $2s=3b$
$\Rightarrow a+b+c=3b$
or $a+c=3b-b=2b$
$\therefore a,b,c$ are in A.P.
And in $△ABC,$
$\cos A+\cos B+\cos C$
$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+\cos C$ by transformation angle formula
$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\left(\frac{C}{2}\right)$ using multiple angle formula
$=1+2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\left(\frac{C}{2}\right)$ since $A+B+C=\pi$
$=1+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\left(\frac{C}{2}\right)$ since $\cos (\frac{\pi }{2}-\theta )=\sin \theta$
$=1+2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right)]$ by taking common terms
$=1+2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2})]$since $A+B+C=\pi$
$=1+2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]$ since $\sin (\frac{\pi }{2}-\theta )=\cos \theta$
$=1+4\sin \left(\frac{C}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{A}{2}\right)$
$=1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$