If $A + B = 60 °$ then $\cos^{2} A + \cos B^{2} - \cos A \cos B =$

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\frac{3}{4}

\frac{4}{3}

\frac{1}{4}

\frac{2}{3}

answer is C.

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Detailed Solution

Given

A + B = 60 °

$\Rightarrow \cos [A + B] = \cos 60$

$\Rightarrow \cos A . \cos B - \sin A . \sin B = \frac{1}{2}$

$\Rightarrow 2 \cos A . \cos B - 1 = 2 \sin A . \sin B$

Square on both sides

$\Rightarrow 4 \cos^{2} A . \cos^{2} B + 1 - 4 \cos A . \cos B$ $= 4 \sin^{2} A . \sin^{2} B$

$\Rightarrow 4 \cos^{2} A . \cos^{2} B - 4 \cos A . \cos B + 1$ $= 4 [1 - \cos^{2} A] [1 - \cos^{2} B]$

$\Rightarrow 4 \cos^{2} A . \cos^{2} B - 4 \cos A . \cos B + 1$ $= 4 [1 - \cos^{2} B - \cos^{2} A + \cos^{2} A . \cos^{2} B]$

$\Rightarrow 4 \cos^{2} A . \cos^{2} B - 4 \cos A . \cos B + 1 = 4 - 4 \cos^{2} B - 4 \cos^{2} A + 4 \cos^{2} A . \cos^{2} B$

$\Rightarrow 4 \cos^{2} B + 4 \cos^{2} A - 4 \cos A . \cos B = 3$

$∴ \cos^{2} B + \cos^{2} A - \cos A . \cos B = \frac{3}{4}$

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