Question

The sides of a triangle are in $A.P.$ If the angles $A$ and $C$ are the greatest and smallest angle respectively, then $4(1-\cos A)(1-\cos C)$ is equal to

• A. $\cos A-\cos C$
• B. $2\cos A\cdot \cos C$
• C. $\cos A+\cos C$
• D. $2\cos A-\cos C$

Answer 1

The correct option is C $\cos A+\cos C$

We have, $2b=a+c$
$\therefore 2\sin B=\sin A+\sin C\Rightarrow 4\sin \frac{B}{2}\cos \frac{B}{2}=2\sin \frac{A+C}{2}\cos \frac{A-C}{2}=2\cos \frac{B}{2}\cos \frac{A-C}{2}\therefore 2\sin \frac{B}{2}=\cos \frac{A-C}{2}\Rightarrow 2\cos \frac{A+C}{2}=\cos \frac{A-C}{2}\cdots \left(i\right)$
Now, $\cos A+\cos C=2\cos \frac{A+C}{2}\cdot \cos \frac{A-C}{2}=2\cos \frac{A+C}{2}\cdot 2\cos \frac{A+C}{2}$
[using equation $\left(i\right)$]
$=4{\cos }^2\frac{A+C}{2}\cdots \left(ii\right)$
$\Rightarrow 4(1-\cos A)(1-\cos C)=4\cdot 2{\sin }^2\frac{A}{2}\cdot 2{\sin }^2\frac{C}{2}=4{\left(2\sin \frac{A}{2}\sin \frac{C}{2}\right)}^2=4{(\cos \frac{A-C}{2}-\cos \frac{A+C}{2})}^2=4{\cos }^2\frac{A+C}{2}\cdots \left(iii\right)$
[using equation $\left(i\right)$]
From $\left(ii\right)$ and $\left(iii\right),$ we get,
$4(1-\cos A)(1-\cos C)=\cos A+\cos C$