MCQ
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$ is equal to :
- ✓$\sin60^\circ$
- B$\cos60^\circ$
- C$\tan60^\circ$
- D$\sin30^\circ$
✓
Answer
Correct option: A.
$\sin60^\circ$
We have to find the value of the following expression
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt3}}{1+\Big(\frac{1}{\sqrt3}\Big)^2}$
$=\frac{\frac{2}{\sqrt3}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt3}}{\frac{4}{3}}$
$\begin{bmatrix}\text{Since}\tan60^\circ=\frac{\sqrt3}{2}\\\text{Since}\tan30^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{\sqrt3}{2}$
$=\sin60^\circ$
Hence the correct option is $(a)$
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$\frac{2\tan30^\circ}{1+\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt3}}{1+\Big(\frac{1}{\sqrt3}\Big)^2}$
$=\frac{\frac{2}{\sqrt3}}{1+\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt3}}{\frac{4}{3}}$
$\begin{bmatrix}\text{Since}\tan60^\circ=\frac{\sqrt3}{2}\\\text{Since}\tan30^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{\sqrt3}{2}$
$=\sin60^\circ$
Hence the correct option is $(a)$
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