Evaluate the following: 30^ 45^ + 45^ 60^ - 60^ 45^ - 30^ 90^

Question

Evaluate the following:
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$

✓

Answer

We have
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}\ \dots(1)$
Now,
$\sin45^\circ=\frac{1}{\sqrt{2}},\ \sin30^\circ=\frac{1}{2},\ \sin90^\circ=1,$ $\tan45^\circ=\cot45^\circ=1,\ \sin60^\circ=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\sec60^\circ=2$
So by substituting above values in equation (1)
We get,
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$ =\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}+\frac{1}{2}-\frac{\frac{\sqrt{3}}{2}}{1}-\frac{\frac{\sqrt{3}}{2}}{1}$
Now by further simplifying
We get,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ} $
$ =\frac{1}{\sqrt{2}\times\sqrt{2}}\times\frac{\sqrt{2}}{1}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
Now, one $\sqrt{2}$ gets cancelled and
We get,
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$ =\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
Now, by taking LCM
We get,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$=\frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
$=\frac{\sqrt{2}}{2}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
$ =\frac{\sqrt{2}+1-\sqrt{3}-\sqrt{3}}{2}$
$=\frac{\sqrt{2}+1-2\sqrt{3}}{2}$
Therefore,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$ $=\frac{\sqrt{2}+1-2\sqrt{3}}{2}$