If ^ - 1 x + 2 ^ - 1 x = 2 3 , then x = - Vidyadip

MCQ

If ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3},$ then $x =$

A
$\sqrt 2 $
B
$3$
✓
$\sqrt 3 $
D
$\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$

✓

Answer

Correct option: C.

$\sqrt 3 $

c
(c) The given equation may be written as

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$

==> ${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$ = $\frac{\pi }{6}$

==> $x = \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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 2. inverse trigonometric function→STD 12 - 2. inverse trigonometric function — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

$a=i+j+k,\,b=2i-4k,\,c=i+\lambda \,j+3k$ are coplanar, then the value of $\lambda $ is

If $\overrightarrow{ a }$ and $\overrightarrow{ b }$ are unit vectors, then the greatest value of $\sqrt{3}|\overrightarrow{ a }+\overrightarrow{ b }|+|\overrightarrow{ a }-\overrightarrow{ b }|$ is

If $A$ and $B$ are two events such that $\text{A}\neq\phi,\text{B}=\phi,$ then,

The position vectors of the points $ A, B, C $ are $(2i + j - k),$ $(3i - 2j + k)$ and $(i + 4j - 3k)$ respectively. These points

Maximize $Z = 3x + 5y,$ subject to constraints: $\text{x}+4\text{y}\leq24,3\text{x}+\text{y}\leq21,\text{x}+\text{y}\geq9,\text{x}\geq0,\text{y}\geq0.$

The determinant $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&3&6\end{array}\,} \right|$ is not equal to

$\int {} $ $e^{\tan \theta} $ $(\sec \theta - \sin \theta )$ $ d\theta $ equals :

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}, \quad 2 \hat{i}+2 \hat{j}+3 \hat{k} \quad$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denotes the length of the angle bisector $\mathrm{AD}$ of $\angle \mathrm{BAC}$ where $\mathrm{D}$ is on the line segment $\mathrm{BC}$, then $2 l^2$ equals:

Let $\vec{a}=\vec{i}-\alpha \vec{j}+\beta \hat{k}, \vec{b}=3 \hat{i}+\beta \hat{j}-\alpha \hat{k}$ and $\vec{c}=-\alpha \hat{i}-2 \hat{j}+\hat{k}$, where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b}=-1$ and $\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=10$, then $(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}}$ is equal to $.....$

Let $f(x)\, = \left\{ {\begin{array}{*{20}{c}}{x + 1,}&{{\rm{when}}}&{x < 2}\\{2x - 1,}&{{\rm{when}}}&{x \ge 2}\end{array}} \right.\,,\,$ then $f'(2) = $