Q.Find the value of the following: $\tan^{-1}(1) + \cos^{-1} \left( -\frac{1}{2} \right) + \sin^{-1} \left( -\frac{1}{2} \right)$

Concept: Inverse Trigonometric Graphs — each inverse function has a specific principal value range.

Step 1:

$\tan^{-1}(1)$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, where $\tan \frac{\pi}{4} = 1$.

So $\tan^{-1}(1) = \frac{\pi}{4}$.

Step 2:

$\cos^{-1}\left(-\frac{1}{2}\right)$ lies in $[0, \pi]$, where $\cos \frac{2\pi}{3} = -\frac{1}{2}$.

So $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.

Step 3:

$\sin^{-1}\left(-\frac{1}{2}\right)$ lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, where $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$. …

The key to solving this lies in remembering the principal value branches of inverse trigonometric functions. Each inverse function is defined to give a single, unique output (the principal value) within a specific interval. Without this, the expression would be ambiguous — every inverse trig function has infinitely many values.

For $\tan^{-1}x$, the principal value lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

For $\cos^{-1}x$, it lies in $[0, \pi]$.

For $\sin^{-1}x$, it lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Once you fix these ranges, each term becomes a simple angle. Then it's just arithmetic.

1. Evaluate $\tan^{-1}(1)$

   We need an angle $\theta$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan\theta = 1$.

   $\tan\frac{\pi}{4} = 1$, and $\frac{\pi}{4}$ lies inside the principal range.

   So $\tan^{-1}(1) = \frac{\pi}{4}$.

2. Evaluate $\cos^{-1}\left(-\frac12\right)$

   We need an angle $\phi$ in $[0, \pi]$ such that $\cos\phi = -\frac12$.

   Cosine is negative in the second quadrant. The reference angle whose cosine is $\frac12$ is $\frac{\pi}{3}$, so the angle in $[0,\pi]$ with cosine $-\frac12$ is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

   Hence $\cos^{-1}\left(-\frac12\right) = \frac{2\pi}{3}$.

3. Evaluate $\sin^{-1}\left(-\frac12\right)$

   We need an angle $\psi$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin\psi = -\frac12$. …

Method: Evaluating a sum of inverse trig values

Use this whenever an expression adds several inverse trig terms; evaluate each separately, then combine.

Steps

Step 1: Evaluate each term in its own principal branch

Handle $\tan^{-1}$, $\cos^{-1}$ and $\sin^{-1}$ one at a time, each restricted to its correct range: $\tan^{-1}\to\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, $\cos^{-1}\to[0,\pi]$, $\sin^{-1}\to\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Step 2: Respect the sign rule of each function …

Common Mistakes

Mistake 1: Writing $\sin^{-1}\left(-\frac{1}{2}\right) = \frac{5\pi}{6}$ or $\frac{7\pi}{6}$

Why it's wrong: those angles are outside sine's branch $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Correct approach: $\sin^{-1}$ is odd, so $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.

Mistake 2: Treating $\cos^{-1}\left(-\frac{1}{2}\right)$ like an odd function

Why it's wrong: $\cos^{-1}(-a) = \pi - \cos^{-1}(a) = \frac{2\pi}{3}$, not $-\frac{\pi}{3}$. Correct approach: apply the supplement rule for negative cosine arguments before adding. …