what is the exact value of tan⁻¹(-1)?\na $\frac{pi}{3}$\nb $-\frac{pi}{4}$\nc $\frac{pi}{6}$\nd $\frac{pi}{4}$

Explanation:

Step1: Recall the definition of inverse - tangent

The inverse - tangent function, $y = \tan^{-1}(x)$, has a range of $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ and $\tan^{-1}(x)$ is the angle $\theta$ in the range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $\tan\theta=x$.

Step2: Find the angle whose tangent is - 1

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. We want to find $\theta$ such that $\tan\theta = - 1$, i.e., $\frac{\sin\theta}{\cos\theta}=-1$ or $\sin\theta=-\cos\theta$. In the unit - circle, for $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, when $\theta =-\frac{\pi}{4}$, $\sin\left(-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}$ and $\cos\left(-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$, and $\tan\left(-\frac{\pi}{4}\right)=\frac{\sin\left(-\frac{\pi}{4}\right)}{\cos\left(-\frac{\pi}{4}\right)}=-1$.

Answer:

B. $-\frac{\pi}{4}$