Question

Evaluate: $\frac{\text{sin}{30}^{\circ }+\text{tan}{45}^{\circ }-\text{cosec}{60}^{\circ }}{\text{cot}{45}^{\circ }+\text{cos}{60}^{\circ }-\text{sec}{30}^{\circ }}$.

Answer 1

We know that,
$\sin {30}^{\circ }=\frac{1}{2},\tan {45}^{\circ }=1,\text{cosec}{60}^{\circ }=\frac{2}{\sqrt{3}}$
and
$\cot {45}^{\circ }=1,\cos {60}^{\circ }=\frac{1}{2},\sec {30}^{\circ }=\frac{2}{\sqrt{3}}$
$\therefore \frac{\sin {30}^{\circ }+\tan {45}^{\circ }-\text{cosec}{60}^{\circ }}{\cot {45}^{\circ }+\cos {60}^{\circ }-\sec {30}^{\circ }}$
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{1+\frac{1}{2}-\frac{2}{\sqrt{3}}}$

$=\frac{1+\frac{1}{2}-\frac{2}{\sqrt{3}}}{1+\frac{1}{2}-\frac{2}{\sqrt{3}}}=1$
$\therefore \frac{\sin {30}^{\circ }+\tan {45}^{\circ }-\text{cosec}{60}^{\circ }}{\cot {45}^{\circ }+\cos {60}^{\circ }-\sec {30}^{\circ }}=1$