Evaluate each of the following

Solution:

We have to find the following expression

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

Now,

$\sin 30^{\circ}=\frac{1}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \tan 60^{\circ}=\sqrt{3}$

So by substituting above values in equation (1)

We get,

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

$=2 \times\left(\frac{1}{2}\right)^{2}-3 \times\left(\frac{1}{(\sqrt{2})}\right)^{2}+(\sqrt{3})^{2}$

$=2 \times \frac{1^{2}}{2^{2}}-3 \times \frac{1^{2}}{(\sqrt{2})^{2}}+3$

$=\frac{2}{4}-\frac{3}{2}+3$

In the above equation the first term $\frac{2}{4}$ gets reduced to $\frac{1}{2}$

Therefore,

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

$=\frac{1}{2}-\frac{3}{2}+3$

$=\frac{1-3}{2}+3$

$=\frac{-2}{2}+3$

In the above equation the first term $\frac{-2}{2}$ gets reduced to $\frac{-1}{1}=-1$

Therefore,

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

$=-1+3$

$=2$

Therefore,

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