left|begin{array}{cc}sin frac{2 pi}{9} & cos | vedclass

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Question

(B) The given determinant is $D = \sin \frac{2 \pi}{9} \cos \frac{5 \pi}{18} - \cos \frac{2 \pi}{9} \sin \frac{5 \pi}{18}$.Using the trigonometric ide

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$-\sin \frac{\pi}{18}$

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(B) The given determinant is $D = \sin \frac{2 \pi}{9} \cos \frac{5 \pi}{18} - \cos \frac{2 \pi}{9} \sin \frac{5 \pi}{18}$.Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we have:$D = \sin \left( \frac{2 \pi}{9} - \frac{5 \pi}{18} ight)$.To subtract the fractions,find a common denominator,which is $18$:$\frac{2 \pi}{9} = \frac{4 \pi}{18}$.So,$D = \sin \left( \frac{4 \pi}{18} - \frac{5 \pi}{18} ight) = \sin \left( -\frac{\pi}{18} ight)$.Since $\sin(-	heta) = -\sin 	heta$,we get $D = -\sin \frac{\pi}{18}$.Thus,the correct option is $B$.

$\left|\begin{array}{cc}\sin \frac{2 \pi}{9} & \cos \frac{2 \pi}{9} \\ \sin \frac{5 \pi}{18} & \cos \frac{5 \pi}{18}\end{array}\right|=$ . . . . . . .

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