Question

Evaluate:
$\frac{\sin {30}^{\circ }}{\sin {45}^{\circ }}+\frac{\tan {45}^{\circ }}{\sec {60}^{\circ }}-\frac{\sin {60}^{\circ }}{\cot {45}^{\circ }}-\frac{\cos {30}^{\circ }}{\sin {90}^{\circ }}$.

Answer 1

We know that,

$\sin {30}^{\circ }=\frac{1}{2},\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\sin {90}^{\circ }=1$

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2},\cos {45}^{\circ }=\frac{1}{\sqrt{2}},\cos {60}^{\circ }=\frac{1}{2}$

$\sec {60}^{\circ }=2,\tan {45}^{\circ }=1$ and $\cot {45}^{\circ }=1$

Now,

$\frac{\sin {30}^{\circ }}{\sin {45}^{\circ }}+\frac{\tan {45}^{\circ }}{\sec {60}^{\circ }}-\frac{\sin {60}^{\circ }}{\cot {45}^{\circ }}-\frac{\cos {30}^{\circ }}{\sin {90}^{\circ }}$

$=\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{\sqrt{2}}\right)}+\frac{1}{2}-\frac{\frac{\sqrt{3}}{2}}{1}-\frac{\frac{\sqrt{3}}{2}}{1}$

$=\frac{\sqrt{2}}{2}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{2}+1-2\sqrt{3}}{2}$