Question

Find the value of the following:
$\frac{5{\cos }^2{60}^o+4{\sec }^2{30}^o-{\tan }^2{45}^o}{{\sin }^2{30}^o+{\cos }^2{30}^o}$

Answer 1

We know that the values of the trignometric functions:

$\sin {30}^0=\cos {60}^0=\frac{1}{2}$,

$\tan {45}^0=1,$ $\cos {30}^0=\frac{\sqrt{3}}{2}$ and $\sec {30}^0=\frac{2}{\sqrt{3}}$

Therefore, the value of $\frac{5{\cos }^2{60}^0+4{\sec }^2{30}^0-{\tan }^2{45}^0}{{\sin }^2{30}^0+{\cos }^2{30}^0}$ is:

$\frac{5{\cos }^2{60}^0+4{\sec }^2{30}^0-{\tan }^2{45}^0}{{\sin }^2{30}^0+{\cos }^2{30}^0}=\frac{\left({5\times \left(\frac{1}{2}\right)}^2\right)+\left({4\times \left(\frac{2}{\sqrt{3}}\right)}^2\right)-1^2}{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{\left(5\times \frac{1}{4}\right)+\left(4\times \frac{4}{3}\right)-1}{\frac{1}{4}+\frac{3}{4}}$ $=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{4}{4}}=\frac{\frac{15+64-12}{12}}{1}=\frac{67}{12}$

Hence, $\frac{5{\cos }^2{60}^0+4{\sec }^2{30}^0-{\tan }^2{45}^0}{{\sin }^2{30}^0+{\cos }^2{30}^0}=\frac{67}{12}$.