MCQ
Find the principal value of $\tan ^{-1}(-\sqrt{3})$
- A$\frac{\pi}{3}$
- ✓$-\frac{\pi}{3}$
- C$-\frac{\pi}{6}$
- D$\frac{5\pi}{6}$
Then, $\tan y=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan \left(-\frac{\pi}{3}\right)$
We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2} \frac{\pi}{2}\right)$ and $\tan \left(-\frac{\pi}{3}\right)$ is $-\sqrt{3}$
Therefore, known that the principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$
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$f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2$
$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$
where $P(x)$ is a polynomial such that $P^{\prime \prime}(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to $.....$
