If 2left|begin{array}{ll}sin ( A + B ) & cos | vedclass

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(B) Given the determinant equation: $2[\sin(A+B)\sin(A-B) - \cos(A+B)\cos(A-B)] + \sqrt{3} = 0$ Using the trigonometric identity $\cos(x+y) = \cos x \

Vedclass · Accepted Answer

$\frac{\pi}{12}$

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(B) Given the determinant equation: $2[\sin(A+B)\sin(A-B) - \cos(A+B)\cos(A-B)] + \sqrt{3} = 0$ Using the trigonometric identity $\cos(x+y) = \cos x \cos y - \sin x \sin y$,we can rewrite the expression inside the brackets as: $-(\cos(A+B)\cos(A-B) - \sin(A+B)\sin(A-B)) = -\cos((A+B) + (A-B)) = -\cos(2A)$ Substituting this into the equation: $2[-\cos(2A)] + \sqrt{3} = 0$ $-2\cos(2A) = -\sqrt{3}$ $\cos(2A) = \frac{\sqrt{3}}{2}$ Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have: $2A = \frac{\pi}{6}$ $A = \frac{\pi}{12}$

If $2\left|\begin{array}{ll}\sin ( A + B ) & \cos ( A + B ) \\ \cos ( A - B ) & \sin ( A - B )\end{array}\right|+\sqrt{3}= 0$,then $A =$ . . . . . . .

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