Evaluate $2 {tan}^{2} 45^{o} + {cos}^{2} 30^{o} - {sin}^{2} 60^{o}$

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Solution

Now,
$2 {tan}^{2} 45^{o} + {cos}^{2} 30^{o} - {sin}^{2} 60^{o}$
$= 2 \times 1^{2} + \frac{3}{4} - \frac{3}{4}$ [ Since $tan 45^{o} = 1, cos 30^{o} = sin 60^{o} = \frac{\sqrt{3}}{2}$ ]
$= 2$ .

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