Question
Evaluate the following:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
✓
Answer
We have
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ\ \dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \sin60^\circ=\frac{\sqrt{3}}{2}$
So by substituting above values in equation (1)
We get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2}-\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=\frac{4}{3}+\frac{1}{\frac{\big(\sqrt{3}\big)^2}{2^2}}-\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=\frac{4}{3}+\frac{2^2}{\big(\sqrt{3}\big)^2}-\frac{1}{2}$
$ =\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4\times2}{3\times2}+\frac{4\times2}{3\times2}-\frac{1\times3}{2\times3}$
$ =\frac{8}{6}+\frac{8}{2}-\frac{3}{6}$
$=\frac{8+8-3}{6}$
$ =\frac{16-3}{6}$
$ =\frac{13}{6}$
Hence,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ=\frac{13}{6}$
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ\ \dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \sin60^\circ=\frac{\sqrt{3}}{2}$
So by substituting above values in equation (1)
We get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2}-\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=\frac{4}{3}+\frac{1}{\frac{\big(\sqrt{3}\big)^2}{2^2}}-\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=\frac{4}{3}+\frac{2^2}{\big(\sqrt{3}\big)^2}-\frac{1}{2}$
$ =\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4\times2}{3\times2}+\frac{4\times2}{3\times2}-\frac{1\times3}{2\times3}$
$ =\frac{8}{6}+\frac{8}{2}-\frac{3}{6}$
$=\frac{8+8-3}{6}$
$ =\frac{16-3}{6}$
$ =\frac{13}{6}$
Hence,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ=\frac{13}{6}$
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