2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

On substituting the values of various T-ratios, we get:
 2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

$=2 \times\left(\frac{1}{2}\right)^{2}+3 \times\left(\frac{1}{\sqrt{2}}\right)^{2}-3 \times\left(\frac{1}{2}\right)^{2}+2 \times(0)^{2}$

$=2 \times \frac{1}{4}+3 \times \frac{1}{2}-3 \times \frac{1}{4}+0$

$=\left(\frac{1}{2}+\frac{3}{2}-\frac{3}{4}\right)=\left(\frac{2+6-3}{4}\right)=\frac{5}{4}$