Q) Evaluate $\frac{5cos^2 60 + 4 sec^2 30 - tan ^2 45}{sin^2 30 + cos^2 30}$

Since sin2θ + cos2θ = 1 $\therefore$ $\frac{5cos^2 60 + 4 sec^2 30 - tan ^2 45}{sin^2 30 + cos^2 30}$ = $\frac{5 cos^2 60 + 4 sec^2 30 - tan ^2 45}{1}$ = 5 cos2 60 + 4 sec2 30 - tan2 45 = $5 (\frac{1}{2})^2 + 4(\frac{2}{\sqrt3})^2 - (1)^2$ = $5 (\frac{1}{4}) + 4(\frac{4}{3}) - 1$ = $\frac{5}{4} + \frac{16}{3} - 1$ = $\frac{67}{12}$