Question

Evaluate: $\frac{\sin {60}^0}{{\cos }^2{45}^0}-cot{30}^0+15\cos {90}^0$

Answer 1

Evaluating the value of the given expression:

$$\begin{gathered} \frac{\sin {60}^0}{{\cos }^2{45}^0}-cot{30}^0+15\cos {90}^0\left[\cos 90°=0,cot30°=\sqrt{3},\sin 60°=\frac{\sqrt{3}}{2},\cos 45°=\frac{1}{\sqrt{2}}\right] \\ =\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}^2}-\sqrt{3}+15\times 0 \\ =\frac{\sqrt{3}}{2}\times \frac{2}{1}-\sqrt{3}+0 \\ =\sqrt{3}-\sqrt{3} \\ =0 \end{gathered}$$

Hence, $\frac{\sin {60}^0}{{\cos }^2{45}^0}-cot{30}^0+15\cos {90}^0$ is equal to zero.