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Questions  

If cosα+cosβ=0=sinα+sinβ then cos2α+cos2β is equal to

a
2cos⁡(α+β)
b
-2cos⁡(α+β)
c
3cos⁡(α+β)
d
None of these

detailed solution

Correct option is B

Given cos⁡α+cos⁡β=0 and sin⁡α+sin⁡β=0On squaring and subtracting both the equations, we get (cos⁡α+cos⁡β)2−(sin⁡α+sin⁡β)2=0⇒cos2⁡α+cos2⁡β+2cos⁡αcos⁡β−sin2⁡α+sin2⁡β+2sin⁡αsin⁡β=0⇒cos2⁡α−sin2⁡α+cos2⁡β−sin2⁡β+2[cos⁡αcos⁡β−sin⁡αsin⁡β]=0⇒cos⁡2α+cos⁡2β+2cos⁡(α+β)=0⇒cos⁡2α+cos⁡2β=−2cos⁡(α+β)

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