Question

\[ \tan^{-1}1+\tan^{-1}2+\tan^{-1}3= \]

• A. \(\frac{3\pi}{4}\)
• B. \(\frac{\pi}{2}\)
• C. \(\pi\)
• D. \(2\pi\)

Hint:

While adding inverse tangent terms, always check the sign of \(1-ab\) to decide the correct quadrant.

Answer 1

The Correct Option is C

Concept: Use the identity: \[ \tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right) \]

Step 1: Let \[ A=\tan^{-1}1+\tan^{-1}2 \] \[ A=\tan^{-1}\left(\frac{1+2}{1-1\cdot 2}\right) \] \[ A=\tan^{-1}\left(\frac{3}{-1}\right) \] Since the denominator is negative, the angle lies in the second quadrant: \[ A=\pi-\tan^{-1}3 \]

Step 2: Add the third term. \[ \tan^{-1}1+\tan^{-1}2+\tan^{-1}3 = \pi-\tan^{-1}3+\tan^{-1}3 \] \[ =\pi \] Therefore, \[ \boxed{\pi} \]