^ -1 (1 3 )-cosec^ -1 (-2)= _________. - Vidyadip

MCQ

$\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\operatorname{cosec}^{-1}(-\sqrt{2})=$ _________.

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

✓

Answer

SELF

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 Model Paper 2→Model Paper 2 — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Given $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{dx}}{{1 + \sin x + \cos x}}} = ln\, 2$, then the value of the def. integral. $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{\sin \,x}}{{1 + \sin x + \cos x}}} \,dx$ is equal to

If the direction ratios of two lines are given by $3lm - 4\,ln + mn = 0$ and $l + 2m + 3n = 0$, then the angle between the lines is

If $ ABCDEF$ is a regular hexagon and $\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} = \lambda \,\overrightarrow {AD} ,$ then $\lambda = $

The area of the region for which $0 < y < 3 - 2x - x^2\, \& \,x > 0$ is :

$\frac{\text{d}}{\text{dx}}\Big\{\tan^{-1}\Big(\frac{\cos\text{x}}{1+\sin\text{x}}\Big)\Big\}$ equals:

$\frac{1}{2}$
$-\frac{1}{2}$
$1$
$-1$

The equation to the normal to the curve $\text{y}=\sin\text{x}$ at (0, 0) is:

x = 0
y = 0
x + y = 0
x - y = 0

Which of the following statements is false?

If $f(x) = \left\{ \begin{array}{l}a{x^2} - b,\,\,when\,\,{\rm{\,\, }}0 \le x < 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,2,\;\,when\,\,{\rm{ }}x = 1\\\,\,\,\,\,x + 1,\,\,\,when\,{\rm{ \,\,1}} \,\, < x \le 2\end{array} \right.$ is continuous at $x = 1$, then the most suitable value of $a, b$ are

The position of a point in time $ ‘ t’ $ is given by $x = a + bt - c{t^2}$, $y = at + b{t^2}$. Its acceleration at time $ ‘t’$ is

$\text{Area}\text{ of}\text{ ellips}\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1\text{ is:}$

$4\pi\text{ ab}\text{ sq.}\text{ units}$
$2\pi\text{ ab}\text{ sq.}\text{ units}$
$\pi\text{ ab}\text{ sq.}\text{ units}$
$\frac{\pi\text{ab}}{2}\text{ sq.}\text{ units}$