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, $2 \sin ^{2} 30^{\circ}-3 \cos ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}=2$.