Solve
$cos (A - B) - cos (180 - (A + B) = 0$

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Solution

$cos (A - B) - cos (180 - (A + B)) = - 2 sin (\frac{A - B + 180 - A - B}{2}) sin (\frac{A - B - 180 + A + B}{2})$
$= - 2 sin (90 - B) sin (A - 90) = 2 cos A cos B$
$cos (A - B) - cos (180 - (A + B)) = 0 ⟹ cos A = 0$ or $cos B = 0 ⟹ A = 90$ or $B = 90$

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Solvecos(A−B)−cos(180−(A+B)=0

cos(A−B)−cos(180−(A+B))=−2sin(A−B+180−A−B2)sin(A−B−180+A+B2) =−2sin(90−B)sin(A−90)=2cosAcosBcos(A−B)−cos(180−(A+B))=0⟹cosA=0 or cosB=0⟹A=90 or B=90

Solve
$cos (A - B) - cos (180 - (A + B) = 0$

$cos (A - B) - cos (180 - (A + B)) = - 2 sin (\frac{A - B + 180 - A - B}{2}) sin (\frac{A - B - 180 + A + B}{2})$
$= - 2 sin (90 - B) sin (A - 90) = 2 cos A cos B$
$cos (A - B) - cos (180 - (A + B)) = 0 ⟹ cos A = 0$ or $cos B = 0 ⟹ A = 90$ or $B = 90$