Question

$\frac{4}{{\cot }^{2}30°}+\frac{1}{{\sin }^{2}30°}-2{\cos }^{2}45°-{\sin }^{2}0°$

Answer 1

On substituting the values of various T-ratios, we get:

$$\begin{gathered} \frac{4}{{\cot }^2{30}^{\mathrm{o}}}+\frac{1}{{\sin }^2{30}^{\mathrm{o}}}-2{\cos }^2{45}^{\mathrm{o}}-{\sin }^2{0}^{\mathrm{o}} \\ =\frac{4}{{\left(\sqrt{3}\right)}^2}+\frac{1}{{\left({\displaystyle \frac{1}{2}}\right)}^2}-2\times {\left(\frac{1}{\sqrt{2}}\right)}^2-{\left(0\right)}^2 \\ =\frac{4}{3}+\frac{1}{{\displaystyle \frac{1}{4}}}-2\times \frac{1}{2}-0 \\ =\frac{4}{3}+4-1 \\ =\frac{4}{3}+3=\frac{4+9}{3}=\frac{13}{3} \end{gathered}$$