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If A and B be acute positive angles satisfying $3 \sin^{2} A + 2 \sin^{2} B = 1, 3 \sin 2 A - 2 \sin 2 B = 0$ then

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B=π4−A2

A=π4−2B

B=π2−A4

A=π4−B2

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detailed solution

Correct option is A

sin⁡2B=(3/2)sin⁡2A,cos⁡2B=3sin2⁡A⇒tan⁡2B=cot⁡A⇒tan⁡Atan⁡2B=1⇒A+2B=π/2⇒B=π/4−A/2

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If A and B be acute positive angles satisfying 3sin2⁡A+2sin2⁡B=1,3sin⁡2A−2sin⁡2B=0 then

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Ready to Test Your Skills?

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If A and B be acute positive angles satisfying $3 \sin^{2} A + 2 \sin^{2} B = 1, 3 \sin 2 A - 2 \sin 2 B = 0$ then