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$\cos 28^{0} + \sin 28^{0} = k^{3} then \cos 17^{0} =$

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Correct option is A

We have cos28∘+sin28∘=k3⇒cos28°+cos62°=k3⇒2cos45°cos17°=k3 ∵cosC+cosD=2cosC+D2cosC-D2⇒cos17°=k32

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cos280+sin280=k3 then cos170=

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$\cos 28^{0} + \sin 28^{0} = k^{3} then \cos 17^{0} =$