Question

Find the value of the following:
$\frac{4{\sin }^2{60}^o-{\cos }^2{45}^o}{{\tan }^2{30}^o+{\sin }^2{0}^o}$

Answer 1

We know that the values of the trignometric functions $\sin {60}^0=\frac{\sqrt{3}}{2}$, $\sin 0^0=0$, $\tan {30}^0=\frac{1}{\sqrt{3}}$ and $\cos {45}^0=\frac{1}{\sqrt{2}}$, therefore, the value of $\frac{4{\sin }^2{60}^0-{\cos }^2{45}^0}{{\tan }^2{30}^0+{\sin }^2{0}^0}$ is:

$\frac{4{\sin }^2{60}^0-{\cos }^2{45}^0}{{\tan }^2{30}^0+{\sin }^2{0}^0}=\frac{(4\times {\left(\frac{\sqrt{3}}{2}\right)}^2)-{\left(\frac{1}{\sqrt{2}}\right)}^2}{{\left(\frac{1}{\sqrt{3}}\right)}^2+{\left(0\right)}^2}=\frac{(4\times \frac{3}{4})-\frac{1}{2}}{\frac{1}{3}}=\frac{3-\frac{1}{2}}{\frac{1}{3}}=\frac{\frac{6-1}{2}}{\frac{1}{3}}=\frac{\frac{5}{2}}{\frac{1}{3}}=\frac{5}{2}\times \frac{3}{1}=\frac{15}{2}$

Hence, $\frac{4{\sin }^2{60}^0-{\cos }^2{45}^0}{{\tan }^2{30}^0+{\sin }^2{0}^0}=\frac{15}{2}$.