cos2⁡10∘−cos⁡10∘cos⁡50∘+cos2⁡50∘=

cos2⁡10∘−cos⁡10∘cos⁡50∘+cos2⁡50∘=122cos2⁡10∘−2cos⁡10∘cos⁡50∘+2cos2⁡50∘=121+cos⁡20∘−cos⁡60∘−cos⁡40∘+1+cos⁡100∘=1232+cos⁡20∘−cos⁡40∘+cos⁡100∘=34+12cos⁡20∘−122sin⁡70∘sin⁡30∘=34+12cos⁡20∘−sin⁡70∘=34.

cos2⁡10∘−cos⁡10∘cos⁡50∘+cos2⁡50∘=

cos2⁡10∘−cos⁡10∘cos⁡50∘+cos2⁡50∘=122cos2⁡10∘−2cos⁡10∘cos⁡50∘+2cos2⁡50∘=121+cos⁡20∘−cos⁡60∘−cos⁡40∘+1+cos⁡100∘=1232+cos⁡20∘−cos⁡40∘+cos⁡100∘=34+12cos⁡20∘−122sin⁡70∘sin⁡30∘=34+12cos⁡20∘−sin⁡70∘=34.

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$\cos^{2} 10^{\circ} - \cos 10^{\circ} \cos 50^{\circ} + \cos^{2} 50^{\circ} =$

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$\frac{3}{2} (1 + \cos 20^{\circ})$

3/4

$\frac{3}{4} + \cos 20^{\circ}$

3/2

answer is B.

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Detailed Solution

$\begin{array}{l} \cos^{2} 10^{\circ} - \cos 10^{\circ} \cos 50^{\circ} + \cos^{2} 50^{\circ} = \frac{1}{2} [2 \cos^{2} 10^{\circ} - 2 \cos 10^{\circ} \cos 50^{\circ} + 2 \cos^{2} 50^{\circ}] \\ = \frac{1}{2} [1 + \cos 20^{\circ} - \cos 60^{\circ} - \cos 40^{\circ} + 1 + \cos 100^{\circ}] = \frac{1}{2} [\frac{3}{2} + \cos 20^{\circ} - \cos 40^{\circ} + \cos 100^{\circ}] \\ = [\frac{3}{4} + \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} 2 \sin 70^{\circ} \sin 30^{\circ}] = \frac{3}{4} + \frac{1}{2} [\cos 20^{\circ} - \sin 70^{\circ}] = \frac{3}{4} . \end{array}$