Question

Evaluate $\frac{4{\cot }^2{60}^{\circ }+{\sec }^2{30}^{\circ }-2{\sin }^2{45}^{\circ }}{{\sin }^2{60}^{\circ }+{\cos }^2{45}^{\circ }}$

Answer 1

On subsituting the values of given trigonometric ratios of standard angles, we get :

$\frac{4{\cot }^{2}60°+{\sec }^{2}30°-2{\sin }^{2}45°}{{\sin }^{2}60°+{\cos }^{2}45°}=\frac{4{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{2}{\sqrt{3}}\right)}^2-2{\left(\frac{1}{\sqrt{2}}\right)}^2}{{\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}=\frac{\frac{4}{3}+\frac{4}{3}-1}{\frac{3}{4}+\frac{1}{2}}=\frac{\frac{4+4-3}{3}}{\frac{3+2}{4}}=\frac{5}{3}\times \frac{4}{5}=\frac{4}{3}$

$\therefore \frac{4{\cot }^{2}60°+{\sec }^{2}30°-2{\sin }^{2}45°}{{\sin }^{2}60°+{\cos }^{2}45°}=\frac{4}{3}$