If A+B=π3 and cosA+cosB=1 , then

A+B=60° cosA+cosB=1 2cos(A+B2)cos(A−B2)=1⇒2cos30°.cos(A−B2)=1 ⇒2(32)cos(A−B2)=1 cos(A−B2)=13 cos(A−B)=2cos2(A−B2)−1 =2(13)−1=2−33 cos(A−B)=−13

A+B=60° cosA+cosB=1 2cos(A+B2)cos(A−B2)=1⇒2cos30°.cos(A−B2)=1 ⇒2(32)cos(A−B2)=1 cos(A−B2)=13 cos(A−B)=2cos2(A−B2)−1 =2(13)−1=2−33 cos(A−B)=−13

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If $A + B = \frac{π}{3}$ and $\cos A + \cos B = 1$ , then

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

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

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

| \cos A - \cos B | = \frac{1}{2 \sqrt{3}}

answer is C.

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A + B = 60 °

$\cos A + \cos B = 1$

$2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2}) = 1 \Rightarrow 2 \cos 30 ° . \cos (\frac{A - B}{2}) = 1$

$\Rightarrow 2 (\frac{\sqrt{3}}{2}) \cos (\frac{A - B}{2}) = 1$

$\cos (\frac{A - B}{2}) = \frac{1}{\sqrt{3}}$

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

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

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

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