Find the value of: $2 {tan}^{2} 45^{o} + {cos}^{2} 30 - {sin}^{2} 60^{o}$

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Solution

$2 {tan}^{2} 45^{o} + {cos}^{2} 30^{o} - {sin}^{2} 60^{o}$
= $2 (1)^{2} + (\frac{\sqrt{3}}{2})^{2} - (\frac{\sqrt{3}}{2})^{2}$ (Since, $tan 45^{o} = 1$ , $cos 30^{o} = \frac{\sqrt{3}}{2}$ and $sin 60^{o} = \frac{\sqrt{3}}{2}$ )
= $2$

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