Question

Prove that
$\frac{\sin {30}^o+\cos {45}^o-\tan {60}^o}{\cot {30}^o-\sin {45}^o-\cos {60}^o}=-1$

Answer 1

$\begin{array}{c}\frac{\sin {30}^0+\cos {45}^0-\tan {60}^0}{\cot {30}^0-\sin {45}^0-\cos {60}^0}\\ \Rightarrow \frac{\frac{1}{2}+\frac{1}{\sqrt{2}}-\sqrt{3}}{\sqrt{3}-\frac{1}{\sqrt{2}}-\frac{1}{2}}\\ \Rightarrow \frac{\frac{1}{2}+\frac{1}{\sqrt{2}}-\sqrt{3}}{-\left(\frac{1}{2}+\frac{1}{\sqrt{2}}-3\right)}\\ \Rightarrow -1\end{array}$