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Evaluate each of the following:$ 2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ} $
Given:
\( 2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ} \)
To do:
We have to evaluate \( 2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ} \).
Solution:
We know that,
$\sin 30^{\circ}=\frac{1}{2}$
$\cos 45^{\circ}=\frac{1}{\sqrt2}$
$\tan 60^{\circ}=\sqrt3$
Therefore,$ 2 \sin ^{2} 30^{\circ}-\cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}=2(\frac{1}{2})^{2} -3(\frac{1}{\sqrt{2}})^{2} +(\sqrt{3})^{2}$
$=2(\frac{1}{4}) -3(\frac{1}{2}) +3$
$=\frac{1}{2}-\frac{3}{2}+3$
$=\frac{1-3+3( 2)}{2}$
$=\frac{-2+6}{2}$
$=\frac{4}{2}$
$=2$
Hence, $2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}=2$.
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