Find the value of:
$\frac{sin\ 30^{o}}{cos\ 45^{o}} \ +\ \frac{cot\ 45^{o}}{sec\ 60^{o}} \ -\ \frac{sin\ 60^{o}}{tan\ 45^{o}} \ -\ \frac{cos\ 30^{o}}{sin\ 90^{o}}$

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

To find: Here we have to find the value of $\frac{sin\ 30^{o}}{cos\ 45^{o}} \ +\ \frac{cot\ 45^{o}}{sec\ 60^{o}} \ -\ \frac{sin\ 60^{o}}{tan\ 45^{o}} \ -\ \frac{cos\ 30^{o}}{sin\ 90^{o}}$.

Solution:

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

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

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

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

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

So, the value is  $\frac{\sqrt{2} \ +\ 1\ -\ 2\sqrt{3}}{2}$.