Question

$\frac{5co{s}^2{60}^{\circ }+4se{c}^2{30}^{\circ }-tan{t}^2{45}^{\circ }}{si{n}^2{30}^{\circ }+co{s}^2{30}^{\circ }}$

Answer 1

$\frac{5co{s}^2{60}^{\circ }+4se{c}^2{30}^{\circ }-tan{t}^2{45}^{\circ }}{si{n}^2{30}^{\circ }+co{s}^2{30}^{\circ }}$

$=\frac{5\times (\frac{1}{2}{)}^2+4\times (\frac{2}{\sqrt{3}}{)}^2-1}{(\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}$

$=\frac{\frac{5}{4}+4\times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

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

$\frac{15+64-12}{12}=\frac{67}{12}$