MCQ
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$ is equal to:
- A$\cos60^\circ$
- B$\sin60^\circ$
- ✓$\tan60^\circ$
- D$\sin30^\circ$
✓
Answer
Correct option: C.
$\tan60^\circ$
We are asked to find the value of the following
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
We know that $\begin{bmatrix}\tan30^\circ=\frac{1}{\sqrt{3}}\\\tan60^\circ=\sqrt{3}\end{bmatrix}$
$=\frac{3}{\sqrt3}$
$=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\sqrt3$
$=\tan60^\circ$
Hence the correct option is (c)
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
We know that $\begin{bmatrix}\tan30^\circ=\frac{1}{\sqrt{3}}\\\tan60^\circ=\sqrt{3}\end{bmatrix}$
$=\frac{3}{\sqrt3}$
$=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\sqrt3$
$=\tan60^\circ$
Hence the correct option is (c)
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