(cos 0° + cos 45° + sin 30°) (sin 90° – cos 45° + cos 60°)

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Solution

As we know that,
$\cos 0 ° = 1 \cos 45 ° = \frac{1}{\sqrt{2}} \sin 30 ° = \frac{1}{2} \sin 90 ° = 1 \cos 60 ° = \frac{1}{2} By substituting these values, we get (\cos 0 ° + \cos 45 ° + \sin 30 °) (\sin 90 ° - \cos 45 ° + \cos 60 °) = (1 + \frac{1}{\sqrt{2}} + \frac{1}{2}) (1 - \frac{1}{\sqrt{2}} + \frac{1}{2}) = ((1 + \frac{1}{2}) + (\frac{1}{\sqrt{2}})) ((1 + \frac{1}{2}) - (\frac{1}{\sqrt{2}})) = {(1 + \frac{1}{2})}^{2} - {(\frac{1}{\sqrt{2}})}^{2} = 1 + \frac{1}{4} + 2 (1) (\frac{1}{2}) - \frac{1}{2} = 1 + \frac{1}{4} + 1 - \frac{1}{2} = \frac{4 + 1 + 4 - 2}{4} = \frac{7}{4} Hence, (\cos 0 ° + \cos 45 ° + \sin 30 °) (\sin 90 ° - \cos 45 ° + \cos 60 °) = \frac{7}{4} .$

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