The principal solution of ^ -1 ( (7 6 ) ) is - Vidyadip

MCQ

The principal solution of $\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ is

A
$\frac{7 \pi}{6}$
✓
$\frac{\pi}{6}$
C
$\frac{-\pi}{6}$
D
$\frac{-5 \pi}{6}$

✓

Answer

Correct option: B.

$\frac{\pi}{6}$

$\text { (b) : In }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right),$
$\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{6}\right)\right)$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{6}\right)\right)=\frac{\pi}{6}$

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

More "M.C.Q (1 Marks)" from Inverse Trigonometric Functions→Inverse Trigonometric Functions — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\int \frac{x d x}{(x-1)(x-2)}$ equals

Evaluate the following integral$:\ \int\sec^2\text{xdx:}$

The solution of the differential equation $\frac{{dy}}{{dx}} = (a{e^{bx}} + c\cos mx)$ is

If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,

$\int\limits_{\sin \,\theta }^{\cos \,\theta } f $ $( x tan \theta) dx$ is (where $ \theta \neq \frac{n \pi }{2},n\in I$)

If $A$ and $B$ are square matrices of same order and $A^{\prime}$ denotes the transpose of $A$, then

Let $R =\{(a, a), (b, b), (c, c), (a, b)\}$ be a relation on set $A = \{a, b, c\}.$ Then, $R$ is:

The three points $\text{ABC}$ have position vectors $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$ are collinear then $(x, y):$

$\int_{}^{} {\frac{{1 + {x^2}}}{{\sqrt {1 - {x^2}} }}dx = } $

Function $f(x) = |x| - |x - 1|$ is monotonically increasing when :