Home › UP Board Class 12 › mathematics › inversetrigonometric › $\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equals
$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equals
A$\tfrac{7\pi}{6}$
B$-\tfrac{\pi}{6}$
C$\tfrac{\pi}{6}$
D$\tfrac{5\pi}{6}$
Answer & Solution
Correct answer: D. $\tfrac{5\pi}{6}$
$\cos^{-1}$ has range $[0, \pi]$, but $\tfrac{7\pi}{6} > \pi$. Reflect about the x-axis: $\cos\tfrac{7\pi}{6} = \cos\left(2\pi - \tfrac{7\pi}{6}\right) = \cos\tfrac{5\pi}{6} = -\tfrac{\sqrt{3}}{2}$, and $\tfrac{5\pi}{6} \in [0,\pi]$. So the inverse returns $\tfrac{5\pi}{6}$.
Related questions
The value of $ in(\tan^{-1}(3/4))$ is:The value of $\cos^{-1}(-1)$ is:For $x \in [-1, 1]$, the identity $ in^{-1}x + \cos^{-1}x$ equals:The principal value of $ in^{-1}(1/2)$ is:All real $x$ satisfying $ in^{-1}(x) + in^{-1}(1-x) = \cos^{-1}(x)$ areFor $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equals$\displaystyle um_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equals$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $