Find the principal value of ^ -1 (3) - Vidyadip

MCQ

Find the principal value of $\cot ^{-1}(\sqrt{3})$

A
$\frac{2\pi}{3}$
✓
$\frac{\pi}{6}$
C
$\frac{\pi}{2}$
D
$\frac{\pi}{3}$

✓

Answer

Correct option: B.

$\frac{\pi}{6}$

b
Let $\cot ^{-1}(\sqrt{3})=y .$ Then cot $y=\sqrt{3}=\cot \left(\frac{\pi}{6}\right)$

We know that the range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$ and $\cot \left(\frac{\pi}{6}\right)=\sqrt{3}$

Therefore, the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $1,\omega ,{\omega ^2}$ are the cube roots of unity, then their product is

Let $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$ be the solution of the equation $4 x^4+8 x^3-17 x^2-12 x+9=0$ and $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16} m$. Then the value of $\mathrm{m}$ is..........

$\tan \left( {{{90}^o} - {{\cot }^{ - 1}}\frac{1}{3}} \right) = $

If the probability that a randomly chosen $6$-digit number formed by using digits $1$ and $8$ only is a multiple of $21$ is $p$, then $96\;p$ is equal to

If the expansion of ${\left( {{y^2} + \frac{c}{y}} \right)^5}$, the coefficient of $y$ will be

Let $f$ be a real valued function defined by

$f(x) = sin^{-1} \left( {\frac{{\,\,1 - \,\,\left| x \right|}}{3}} \right) + cos^{-1}\left( {\frac{{\left| x \right|\,\, - \,\,3}}{5}} \right)$ .

Then domain of $f(x)$ is given by :

If the ${n^{th}}$ term of a series be $3 + n\,(n - 1)$, then the sum of $n$ terms of the series is

The function ${x^4} - 4x$ is decreasing in the interval

$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $

$\mathop {Lim}\limits_{x\,\, \to \,\,\infty } $ $\frac{{2 + 2x + \sin 2x}}{{(2x + \sin 2x){e^{\sin x}}}}$ is :