Suppose $\sqrt{9 - 8 \cos 40^{\circ}} = a + bsec 40^{\circ}$ ,where a and b are rational numbers. Then $|a + b|$ equals.

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$\frac{1}{2}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \sqrt{9 - 8 \cos 40^{\circ}} = a + bsec 40^{\circ} \\ \Rightarrow \sqrt{9 \cos^{2} 40^{\circ} - 8 \cos^{3} 40^{\circ}} = acos 40^{\circ} + b - - - - (i) \end{array}$

now,

$\begin{array}{l} 9 \cos^{2} 40^{\circ} - 8 \cos^{3} 40^{\circ} = 9 \cos^{2} 40^{\circ} - 2 [3 \cos 40^{\circ} + \cos 120^{\circ}] \\ (∵ \cos 3 θ = - 3 \cos θ + 4 \cos^{3} θ) \\ = 9 \cos^{2} 40^{\circ} - 6 \cos 40^{\circ} + 1 \\ = {(3 \cos 40^{\circ} - 1)}^{2} \end{array}$

$\begin{array}{l} ∴ \sqrt{9 \cos^{2} 40^{\circ} - 8 \cos^{3} 40^{\circ}} = 3 \cos 40^{\circ} - 1 = acos 40^{\circ} + b \\ \Rightarrow a = 3, b = - 1 \\ ∴ a + b = 2 \end{array}$