Question

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

Answer 1

We know that the values of the trignometric functions:

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

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

$\frac{5{\sin }^2{30}^0+{\cos }^2{45}^0-4{\tan }^2{30}^0}{2\sin {30}^0+\cos {30}^0+\tan {45}^0}=\frac{\left({5\times \left(\frac{1}{2}\right)}^2\right)+{\left(\frac{1}{\sqrt{2}}\right)}^2-\left({4\times \left(\frac{1}{\sqrt{3}}\right)}^2\right)}{(2\times \frac{1}{2})+\frac{\sqrt{3}}{2}+1}$

$=\frac{\left(5\times \frac{1}{4}\right)+\frac{1}{2}-\left(4\times \frac{1}{3}\right)}{1+\frac{\sqrt{3}}{2}+1}=\frac{\frac{5}{4}+\frac{1}{2}-\frac{4}{3}}{2+\frac{\sqrt{3}}{2}}=\frac{\frac{15+6-16}{12}}{\frac{4+\sqrt{3}}{2}}=\frac{\frac{5}{12}}{\frac{4+\sqrt{3}}{2}}=\frac{5}{12}\times \frac{2}{4+\sqrt{3}}=\frac{5}{6(4+\sqrt{3})}$

$=\frac{5}{24+6\sqrt{3}}$

Hence, $\frac{5{\sin }^2{30}^0+{\cos }^2{45}^0-4{\tan }^2{30}^0}{2\sin {30}^0+\cos {30}^0+\tan {45}^0}=\frac{5}{24+6\sqrt{3}}$.