Download App

Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions1 Mark
Question
Find the principal values of: $\cot ^{-1}(1)$
✓

Answer

(b) : Let $\cot ^{-1}(1)=\theta \Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4} \in(0, \pi)$
$\therefore$ Principal value of $\cot ^{-1}(1)$ is $\frac{\pi}{4}$.

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 Inverse Trigonometric Functions→Inverse Trigonometric Functions — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $f(x) = 4x^8,$ then:
→
If $\frac{d}{d x}(f(x))=\log x$, then $f(x)$ equals:
→
$\int\frac{\cos2\text{x dx}}{(\sin\text{x}+\cos\text{x})^2}=$
  1. $-\frac{1}{\sin\text{x}+\cos\text{x}}+\text{c}$
  2. $\log|\sin\text{x}+\cos\text{x }|+\text{c}$
  3. $\log|\sin\text{x}-\cos\text{x }|+\text{c}$
  4. $\frac{1}{(\sin\text{x}+\cos\text{x})^2}$
→
The equation of the plane through point (1, 2, -3) which is parallel to the plane 3x - 5y + 2z = 11 is given by:
→
If $f\text{(x)} = \text{(ax^2 + b)}^3,$ then the function $g$ such that $f(g\text{(x)}) = \text{g(f(x))}$ is given by:
→
If $\alpha,\beta,\gamma$ are the angles which a directed line makes with the positive directions of the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:
  1. 1
  2. 4
  3. 3
  4. 2
→
The corner points of the feasible region determined by the following system of linear inequalities: $2\text{x}+\text{y}\leq10,\text{x}+3\text{y}\leq15, \text{x},\text{y}\geq0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px + qy, where p, q > 0. Conditions on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:
→
The vector equation of the plane containing the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$ and the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ is:
  1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
  2. $\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
  3. $\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
  4. $\text{None of these}$
→
If AD, BE and CF are $\triangle\text{ABC},$ then $\vec{\text{AD}}+\vec{\text{BE}}+\vec{\text{CF​}}$
  1. $\vec{0}$
  2. 1
  3. 0
  4. 2
→
If $y^2(2-x)=x^3$, then $\left(\frac{d y}{d x}\right)_{(1,1)}$ is equal to
→