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Q.

sin2⁡π8+A2−sin2⁡π8−A2 is equal to

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a

(1/2)sin⁡A

b

(1/2)cos⁡A

c

2sin⁡A

d

2cos⁡A

answer is A.

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Detailed Solution

Use :sin2⁡A−sin2⁡B=sin⁡(A−B)sin⁡(A+B)
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sin2⁡π8+A2−sin2⁡π8−A2 is equal to