alleryThe principal value of $\tan^{-1}(1)$ is
A$\tfrac{\pi}{3}$
B$\tfrac{\pi}{4}$
C$\tfrac{\pi}{2}$
D$\tfrac{\pi}{6}$
Answer & Solution
Correct answer: B. $\tfrac{\pi}{4}$
$\tan^{-1}$ returns values in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.
We need $\tan\theta = 1$ with $\theta$ in that range, which gives $\theta = \tfrac{\pi}{4}$.
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Related questions
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) = $$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equals$ in^{-1}\left( in\tfrac{2\pi}{3}\right)$ equals$\cot^{-1}\left(-\tfrac{1}{ qrt{3}}\right) = $$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equals