Question

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

Answer 1

We know that,
$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}\Rightarrow {\cos }^2{30}^{\circ }=\frac{3}{2}$
$\cos {60}^{\circ }=\frac{1}{2}\Rightarrow {\cos }^2={60}^{\circ }=\frac{1}{4}$
$\sec {30}^{\circ }=\left(\frac{2}{\sqrt{3}}\right)\Rightarrow {\sec }^2{30}^{\circ }=\frac{4}{3}$
$\tan {45}^{\circ }=1\Rightarrow {\tan }^2{45}^{\circ }=1$
$\sin {30}^{\circ }=\frac{1}{2}\Rightarrow {\sin }^2{30}^{\circ }=\frac{1}{4}$
Now,
$\frac{(5{\cos }^2{60}^{\circ }+4{\sec }^2{30}^{\circ }-{\tan }^2{45}^{\circ })}{({\sin }^2{30}^{\circ }+{\cos }^2{30}^{\circ })}$
$=\frac{5(\frac{1}{2}{)}^2+4(\frac{2}{\sqrt{3}}{)}^2-(1{)}^2}{(\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}$
$=\frac{5\times \frac{1}{4}+4\times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}$
$=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1+3}{4}}$
$=\frac{\frac{15+64-12}{12}}{\frac{4}{4}}$
$=\frac{67}{12}$.