Question

Solve that : $\frac{{5cos}^{2}60+{4sec}^{2}30-{tan}^{2}45}{{sin}^{2}30+{cos}^{2}30}$

Answer 1

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

$\Rightarrow \frac{5{\left(\frac{1}{2}\right)}^2+4{\left(\frac{2}{\sqrt{3}}\right)}^2-(1{)}^2}{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$\Rightarrow \frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

$\Rightarrow \frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{4}{4}}$

$\Rightarrow \frac{3+64}{12}=\frac{67}{12}$.