Question

Prove that $\frac{4}{3}{\sec }^2{60}^{\circ }-cose{c}^2{30}^{\circ }-\frac{3}{4}{\tan }^2{45}^{\circ }-2{\sin }^2{60}^{\circ }=\frac{-11}{12}$

Answer 1

$\frac{4}{3}{\sec }^{2}60-{\csc }^{2}30-\frac{3}{4}{\tan }^{2}45-2{\sin }^{2}60=\frac{-11}{12}LHS=\frac{4}{3}\left(2^2\right)-\left(2^2\right)-\frac{3}{4}{\left(1\right)}^2-2{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{16}{3}-4-\frac{3}{4}-\frac{3}{2}=\frac{128-96-18-36}{24}=\frac{-11}{12}$