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$a^{2} \cos^{2} \frac{2 π}{3} - 4 a^{2} \tan^{2} \frac{3 π}{4} + 2 a^{2} \sin^{2} \frac{2 π}{3} =$

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3a24

−9a24

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detailed solution

Correct option is B

a2−122−4a2−12+2a2322=−9a24

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a2cos22π3−4a2tan23π4+2a2sin22π3=

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$a^{2} \cos^{2} \frac{2 π}{3} - 4 a^{2} \tan^{2} \frac{3 π}{4} + 2 a^{2} \sin^{2} \frac{2 π}{3} =$