Question

Evaluate: \[ \tan^{-1}\left(\frac12\right) +\tan^{-1}\left(\frac13\right) +\tan^{-1}\left(\frac23\right) +\tan^{-1}\left(\frac15\right) \]

• A. \(\dfrac{\pi}{4}\)
• B. \(\tan^{-1}\left(\dfrac{7}{11}\right)\)
• C. \(\dfrac{\pi}{2}\)
• D. \(\tan^{-1}\left(\dfrac{23}{24}\right)\)

Hint:

Look for pairs satisfying \[ \frac{a+b}{1-ab}=1. \] Such pairs immediately simplify to \(\tan^{-1}(1)=\frac{\pi}{4}\).

Answer 1

The Correct Option is C

Concept: The standard formula is \[ \tan^{-1}a+\tan^{-1}b = \tan^{-1} \left( \frac{a+b}{1-ab} \right), \] provided the principal value conditions are satisfied. We shall combine the terms systematically.

Step 1: Combine the first two inverse tangents. Let \[ A= \tan^{-1}\left(\frac12\right) + \tan^{-1}\left(\frac13\right). \] Then \[ A = \tan^{-1} \left( \frac{\frac12+\frac13} {1-\frac16} \right). \] \[ = \tan^{-1} \left( \frac{\frac56} {\frac56} \right). \] \[ = \tan^{-1}(1) = \frac{\pi}{4}. \]

Step 2: Combine the remaining two terms. Let \[ B= \tan^{-1}\left(\frac23\right) + \tan^{-1}\left(\frac15\right). \] Then \[ B = \tan^{-1} \left( \frac{\frac23+\frac15} {1-\frac{2}{15}} \right). \] \[ = \tan^{-1} \left( \frac{\frac{13}{15}} {\frac{13}{15}} \right). \] \[ = \tan^{-1}(1) = \frac{\pi}{4}. \]

Step 3: Add the two results. Therefore, \[ A+B = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}. \] Conclusion: \[ \boxed{\frac{\pi}{2}} \]