Question

Principal value of \(\tan^{-1}(\sqrt3)+\tan^{-1}(1)\) is :

• A. \(\frac{\pi}{12}\)
• B. \(\frac{\pi}{4}\)
• C. \(-\frac{\pi}{12}\)
• D. \(\frac{\pi}{2}\)

Answer 1

The Correct Option is C

To find the principal value of \(\tan^{-1}(\sqrt{3}) + \tan^{-1}(1)\)

We begin by recognizing standard angle values:

• \(\tan^{-1}(1) = \dfrac{\pi}{4}\)
• \(\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}\)

So:

\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = \dfrac{\pi}{3} + \dfrac{\pi}{4} = \dfrac{4\pi + 3\pi}{12} = \dfrac{7\pi}{12} \]

However, this is not in the principal value range of inverse tangent, which is \((-\dfrac{\pi}{2}, \dfrac{\pi}{2})\).

So instead, we apply the identity:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \begin{cases} \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right), & \text{if } ab < 1 \\ \pi + \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right), & \text{if } ab > 1 \end{cases} \]

With \(a = \sqrt{3}, b = 1\), we find:

\[ ab = \sqrt{3} > 1 \Rightarrow \text{Use the second case} \]

So compute:

\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = \pi + \tan^{-1}\left(\dfrac{\sqrt{3} + 1}{1 - \sqrt{3}}\right) \]

Now simplify:

\[ \dfrac{\sqrt{3} + 1}{1 - \sqrt{3}} \cdot \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}} = \dfrac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1 - 3} = \dfrac{(4 + 2\sqrt{3})}{-2} = -2 - \sqrt{3} \]

So we now get:

\[ \tan^{-1}(\sqrt{3}) + \tan^{-1}(1) = \pi + \tan^{-1}(-2 - \sqrt{3}) \]

Now, \(\tan^{-1}(-2 - \sqrt{3})\) is negative, meaning it's in the 4th quadrant. So:

\[ \pi + \tan^{-1}(-2 - \sqrt{3}) = \pi - \theta \]

Where \(\theta = \tan^{-1}(2 + \sqrt{3}) \approx \frac{\pi}{12}\) Thus:

\[ \pi - \frac{\pi}{12} = \frac{12\pi - \pi}{12} = \frac{11\pi}{12} \]

But this is not in the principal range \((-\dfrac{\pi}{2}, \dfrac{\pi}{2})\). Hence, the principal value is:

\[ \boxed{-\frac{\pi}{12}} \]