^ - 1 1 2 + 2 ^ - 1 1 2 is equal to - Vidyadip

MCQ

${\cos ^{ - 1}}\frac{1}{2} + 2{\sin ^{ - 1}}\frac{1}{2}$ is equal to

A
$\frac{\pi }{4}$
B
$\frac{\pi }{6}$
C
$\frac{\pi }{3}$
✓
$\frac{{2\pi }}{3}$

✓

Answer

Correct option: D.

$\frac{{2\pi }}{3}$

d
(d) ${\cos ^{ - 1}}\frac{1}{2} + 2{\sin ^{ - 1}}\frac{1}{2} $

$= \frac{\pi }{3} + \frac{{2\pi }}{6} = \frac{{2\pi }}{3}$.

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

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

Similar questions

$\mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} = $

The triangle made by complex numbers $z_1$ , $z_2$ ,$ - \omega {z_1} - {\omega ^2}{z_2}$ on argand plane is

Suppose $\left| {\begin{array}{*{20}{c}}
{f'\left( x \right)}&{f\left( x \right)} \\
{f''\left( x \right)}&{f'\left( x \right)}
\end{array}} \right| = 0$ where $f(x)$ is continuously differentiable function with $f'(x) \ne 0$ and satisfy $f(0) = 1$ and $f'(0) = 2$ , then the number of solution $(s)$ of equation $f(x) = x^2$ is equal to

The value of the integral $\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ is (where $c$ is a constant of integration)

Let $|\cos \theta \cos (60-\theta) \cos (60-\theta)| \leq \frac{1}{8}, \theta \in[0,2 \pi]$

Then, the sum of all $\theta \in[0,2 \pi]$, where $\cos 3 \theta$ attains its maximum value, is :

In tossing $10$ coins, the probability of getting exactly $5$ heads is

The integrating factor of the differential equation $(x\log x)\frac{{dy}}{{dx}} + y = 2\log x$ is

Suppose that the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ are $\left(f_1, 0\right)$ and $\left(f_2, 0\right)$ where $f_1>0$ and $f_2<0$. Let $P _1$ and $P _2$ be two parabolas with a common vertex at $(0,0)$ and with foci at $\left(f_1, 0\right)$ and $\left(2 f_2, 0\right)$, respectively. Let $T_1$ be a tangent to $P_1$ which passes through $\left(2 f_2, 0\right)$ and $T_2$ be a tangent to $P_2$ which passes through $\left(f_1, 0\right)$. The $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$, then the value of $\left(\frac{1}{m^2}+m_2^2\right)$ is

Let $\mathrm{x}^{\mathrm{k}}+\mathrm{y}^{\mathrm{k}}=\mathrm{a}^{\mathrm{k}},(\mathrm{a}, \mathrm{K}>0)$ and $\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\frac{1}{3}}=0$ then $\mathrm{k}$ is

The circle ${x^2} + {y^2} + 4x - 4y + 4 = 0$ touches