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Q.

If cos⁡α+cos⁡β=0=sin⁡α+sin⁡β, then cos⁡2α+cos⁡2β is equal to

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a

−2sin⁡(α+β)

b

−2cos⁡(α+β)

c

2sin⁡(α+β)

d

2cos⁡(α+β)

answer is B.

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Detailed Solution

(cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2=0 or  cos2⁡α+cos2⁡β+2cos⁡αcos⁡β−sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=0  or  cos⁡2α+cos⁡2β=−2(cos⁡αcos⁡β−sin⁡αsin⁡β)=−2cos⁡(α+β)
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