Evaluate each of the following

Solution:

We have,

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}$....(1)

Now,

$\sin 30^{\circ}=\frac{1}{2}, \sin 90^{\circ}=\cos 0^{\circ}=1, \tan 30^{\circ}=\frac{1}{\sqrt{3}}, \tan 60^{\circ}=\sqrt{3}$

So by substituting above values in equation (1)

We get,

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}$

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

Now, $\sqrt{3}$ present in the denominator of above expression gets cancelled and we get,

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}$

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

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

Now by taking LCM in the above expression we get,

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}$

$=\frac{1}{2}-\frac{1 \times 2}{1 \times 2}+\frac{2 \times 2}{1 \times 2}$

$=\frac{1}{2}-\frac{2}{2}+\frac{4}{2}$

$=\frac{1-2+4}{2}$

$=\frac{5-2}{2}$

$=\frac{3}{2}$

Therefore,

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}=\frac{3}{2}$