Question

$A+B=\frac{\pi }{3};\cos A+\cos B=1$, value of $|\cos A-\cos B|$ is

• A. $\frac{1}{3}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$\sqrt{\frac{2}{3}}$

$\cos A+\cos B=1\to 1,A+B=\frac{\pi }{3}$

$2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=1[\because \cos x+\cos y=2\cos \left(\frac{X+Y}{2}\right)\cos \left(\frac{X-Y}{2}\right)]$

$2\cos \left(\frac{\pi }{6}\right)\cos \left(\frac{A-B}{2}\right)=1$

$\therefore \cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{3}}(\because \cos 30°=\frac{\sqrt{3}}{2})$

$\cos (A-B)=2{\cos }^2\left(\frac{A-B}{2}\right)-1(\because \cos 2\theta =2{\cos }^2\theta -1)$$=2\left(\frac{1}{3}\right)-1$

$\cos (A-B)=\frac{-1}{3}$

$\cos (A+B)+\cos (A-B)=\cos 60-\frac{1}{3}$

$=\frac{1}{2}-\frac{1}{3}$

$2\cos A\cos B=\frac{1}{6}$

$\cos A+\cos B=1$

Squaring on both sides

${(\cos A+\cos B)}^2=1$

${(\cos A-\cos B)}^2+4\cos A\cos B=1$

${(\cos A-\cos B)}^2+\frac{1}{3}=1(\because 2\cos A-\cos B=\frac{1}{6})$

$\therefore |\cos A-\cos B|=\sqrt{\frac{2}{3}}$