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 $(c)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The shadow of a $5 m$ long stick is $2 m$ long. At the same time the length of the shadow of a $12.5 m$ high tree $($in m$)$ is
→In the given figure, $D E \| A B, A D=2 x$, $D C=x+3, B E=2 x-1$ and $C E=x$. Then, find the value of $x$.
→If the point $P(x, y)$ is equidistant from $A(5, 1)$ and $B(-1, 5)$, then
→Consider the following distribution :
The sum of the lower limits of the median class and the modal class is :
→| Class | $0-5$ | $5-10$ | $10-15$ | $15-20$ | $20-25$ |
| Frequency | $10$ | $15$ | $12$ | $20$ | $9$ |
If angles $A, B, C $ to a $\triangle\text{ABC}$ from an increasing $AP,$ then $\sin B =$
→If the shadow of a boy $'x\ ’$ metres high is $1.6m$ and the angle of elevation of the sun is $45^\circ,$ then the value of $'x\ ’$ is :
→If $\triangle O A Q \sim \triangle O B P, P B$ and $Q A$ are perpendiculars to segment $A B$ and $P O=5 \ cm$, $Q O=6 \ cm$ and $\operatorname{ar}(\triangle P O B)=125 \ cm ^2$, then the area of $\triangle Q O A$ is
→The probability that a non-leap year has 53 Sundays, is
The $\text{HCF}$ of $256,442$ and $940$ is :
→The ratio of the sum and product of the roots of the equation $7x^2- 12x + 18 = 0$ is :
→