Question
Evaluate the following:
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
✓
Answer
We have to find the following expression
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ\dots(1)$
Now,
$\tan30^\circ=\frac{1}{\sqrt3},\ \tan60^\circ=\sqrt3,\ \tan45^\circ=1$
So by substituting above values in equation (1)
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\Big(\frac{1}{\sqrt3}\Big)^2+(\sqrt3)^2+(1)^2$
$=\frac{1^2}{(\sqrt3)^2}+(\sqrt3)^2+1$
$=\frac{1}{3}+3+1$
$=\frac{1}{3}+4$
Now by taking LCM
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\frac{1}{3}+\frac{4\times3}{1\times3}$
$=\frac{1}{3}+\frac{12}{3}$
$=\frac{1+12}{3}$
$=\frac{13}{3}$
Therefore,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ=\frac{13}{3}$
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ\dots(1)$
Now,
$\tan30^\circ=\frac{1}{\sqrt3},\ \tan60^\circ=\sqrt3,\ \tan45^\circ=1$
So by substituting above values in equation (1)
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\Big(\frac{1}{\sqrt3}\Big)^2+(\sqrt3)^2+(1)^2$
$=\frac{1^2}{(\sqrt3)^2}+(\sqrt3)^2+1$
$=\frac{1}{3}+3+1$
$=\frac{1}{3}+4$
Now by taking LCM
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\frac{1}{3}+\frac{4\times3}{1\times3}$
$=\frac{1}{3}+\frac{12}{3}$
$=\frac{1+12}{3}$
$=\frac{13}{3}$
Therefore,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ=\frac{13}{3}$
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