cot^{-1}(-sqrt{3}) = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that for the inverse cotangent function,the range is $(0, \pi)$.Since $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$,we have:$\cot^{-1}(

Vedclass · Accepted Answer

$\frac{5\pi}{6}$

Vedclass · Answer

(B) We know that for the inverse cotangent function,the range is $(0, \pi)$.Since $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$,we have:$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$We know that $\cot(\frac{\pi}{6}) = \sqrt{3}$,so $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.Therefore,$\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

$\cot^{-1}(-\sqrt{3}) = $

Similar Questions

$\sin (\cot ^{ - 1}x) = $

Find the principal value of $cosec^{-1}(-\sqrt{2})$.

Which of the following graphs is the graph of $f(x) = \tan^{-1} x, (x \in R)$?

$\cos \left(\tan^{-1} x\right) = . . . . . . . (|x| < 1)$.

Write the function in the simplest form: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}, |x| < a$

Vedclass Test Series

Exam Paper Generator

Online Exam Module