Q.Find the value of $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$.

Principal Value of Inverse Sine

The equation $\sin\theta = x$ has infinitely many solutions. If $\sin\theta = \tfrac12$, then $\theta$ could be $\tfrac{\pi}{6}$, $\tfrac{5\pi}{6}$, $\tfrac{13\pi}{6}$, and so on. To make $\sin^{-1}$ a genuine function, we must agree on one answer. That agreed-upon answer is called the principal value.

Restricting the range

Sine is one-to-one on $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$, and on this interval it climbs through every value from $-1$ to $1$ exactly once. So we define:

• Domain: $x \in [-1, 1]$ (sine never exceeds these values).
• Principal value range: $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$.

The principal value is the unique angle in this closed interval whose sine is $x$.

Reading off values

• $\sin^{-1}\left(\tfrac12\right) = \tfrac{\pi}{6}$, since $\tfrac{\pi}{6} \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ and $\sin\tfrac{\pi}{6} = \tfrac12$.
• $\sin^{-1}\left(-\tfrac12\right) = -\tfrac{\pi}{6}$ — the answer can be negative, because the range dips to $-\tfrac{\pi}{2}$.
• $\sin^{-1}(1) = \tfrac{\pi}{2}$ and $\sin^{-1}(0) = 0$.

The classic trap: $\sin^{-1}(\sin x)$

Many students write $\sin^{-1}(\sin x) = x$ automatically. This is true only when $x$ already lies in $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$. Otherwise you must return the principal value — the equivalent angle inside the range. …