MCQ
$\tan 15^\circ = $
  • A
    $\frac{1}{3}$
  • B
    $\sqrt 3 - 2$
  • $2 - \sqrt 3 $
  • D
    None of these

Answer

Correct option: C.
$2 - \sqrt 3 $
c
(c) $\tan {15^o} = \tan ({45^o} - {30^o})$

$ = \frac{{1 - 1/\sqrt 3 }}{{1 + 1/\sqrt 3 }} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}} $

$= 2 - \sqrt 3 $.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ is equal to:
If the vectors $2i + j - k,\, - i + 2j + \lambda k$ and $ - 5i + 2j - k$ are coplanar, then the value of $\lambda $ is equal
Let $A =\left(\begin{array}{cc} m & n \\ p & q \end{array}\right), d =| A | \neq 0| A - d (\operatorname{Adj} A )|=0$. Then
The vector  $ c$  directed along the internal bisector of the angle between the vectors $a = 7i - 4j - 4k$ and $b = - 2i - j + 2k$ with $|c|\, = 5\sqrt 6 ,$ is
Let $a, b$, c be three distinct real numbers, none equal to one. If the vectors $a \hat{i}+\hat{j}+\hat{k}, \hat{i}+b \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+ c \hat{k}$ are coplanar, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ is equal to
Let $a$ and $b$ be roots of ${x^2} - 3x + p = 0$ and let $c$ and $d$ be the roots of ${x^2} - 12x + q = 0$, where $a,\;b,\;c,\;d$ form an increasing G.P. Then the ratio of $(q + p):(q - p)$ is equal to
If $i = \sqrt { - 1} $, then $1 + {i^2} + {i^3} - {i^6} + {i^8}$ is equal to
Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x)=\frac{\{x\}}{1+[x]^2}$, where $[x]$ is the greatest integer less than or equal to $x$, and $\left\{x{\}}=x-[x]\right.$. Which of the following statements are true?

$I.$ The range of $f$ is a closed interval.

$II.$ $f$ is continuous on $R$.

$III.$ $f$ is one-one on $R$

If $a\,.i\,=a\,.\,(i+j)=a\,.\,(i+j+k)$ , then $a = $
If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$, and $x=1$. Then

$(A)$ $S \geq \frac{1}{ e }$ $(B)$ $S \geq 1-\frac{1}{ e }$

$(C)$ $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$ $(D)$ $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$