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

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Solution

We know that the values of the trignometric functions $sin 30^{0} = \frac{1}{2}$ , $cos 60^{0} = \frac{1}{2}$ and $tan 45^{0} = 1$ , therefore, the value of $sin 30^{0} cos 60^{0} - {tan}^{2} 45^{0}$ is:

$sin 30^{0} cos 60^{0} - {tan}^{2} 45^{0} = (\frac{1}{2} \times \frac{1}{2}) - 1^{2} = \frac{1}{4} - 1 = \frac{1 - 4}{4} = - \frac{3}{4}$

Hence, $sin 30^{0} cos 60^{0} - {tan}^{2} 45^{0} = - \frac{3}{4}$ .

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