Question
Evaluate the following:
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
✓
Answer
We have,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}\ \dots(1)$
Now,
$\sin30^\circ=\frac{1}{2},\ \sin90^\circ=\cos0^\circ=1,$ $\tan30^\circ=\frac{1}{\sqrt{3}},\tan60^\circ=\sqrt{3}$
So by substituting above values in equation (1)
We get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2\times1}{\frac{1}{\sqrt{3}}\times\sqrt{3}}$
Now, $\sqrt{3}$ present in the denominator of above expression gets cancelled and we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2}{1}$
$=\frac{1}{2}-1+2$
Now by taking LCM in the above expression we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{1}{2}-\frac{1\times2}{1\times2}+\frac{2\times2}{1\times2}$
$=\frac{1}{2}-\frac{2}{2}+\frac{4}{2}$
$=\frac{1-2+4}{2}$
$=\frac{5-2}{2}$
$=\frac{3}{2}$
Therefore,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}=\frac{3}{2}$
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}\ \dots(1)$
Now,
$\sin30^\circ=\frac{1}{2},\ \sin90^\circ=\cos0^\circ=1,$ $\tan30^\circ=\frac{1}{\sqrt{3}},\tan60^\circ=\sqrt{3}$
So by substituting above values in equation (1)
We get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2\times1}{\frac{1}{\sqrt{3}}\times\sqrt{3}}$
Now, $\sqrt{3}$ present in the denominator of above expression gets cancelled and we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2}{1}$
$=\frac{1}{2}-1+2$
Now by taking LCM in the above expression we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{1}{2}-\frac{1\times2}{1\times2}+\frac{2\times2}{1\times2}$
$=\frac{1}{2}-\frac{2}{2}+\frac{4}{2}$
$=\frac{1-2+4}{2}$
$=\frac{5-2}{2}$
$=\frac{3}{2}$
Therefore,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}=\frac{3}{2}$
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