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$If \sqrt{9 - 8 \cos 40^{\circ}} = a + b \sec 40^{\circ} a, b \in I, value of a + b is$

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Correct option is B

9−8cos⁡40∘=sec⁡40∘9cos2⁡40∘−8cos3⁡40∘=sec⁡40∘9cos2⁡40∘-2cos⁡120∘−6cos⁡40∘ since 4 cos3θ =cos 3θ +3 cos θ=1cos⁡40∘9cos2⁡40∘+1−6cos⁡40∘=1cos⁡40∘(3cos⁡400−1)2=1cos⁡40∘(3cos⁡40o−1)=3−sec⁡40∘=a+bsec⁡40∘ (given) ∴a=3,b=−1 Hence a+b=3−1=2

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