Question

The value of \( \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) \) is equal to

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( 0 \)
• D. \( \pi \)

Hint:

Always be mindful of the conditions under which the inverse tangent sum formula is applied, especially when \( xy>1 \). The range of \( \tan^{-1}(x) \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which helps in determining the correct quadrant for the sum.

Answer 1

The Correct Option is D

Step 1: Evaluate \( \tan^{-1}(1) \).
We know that \( \tan(\frac{\pi}{4}) = 1 \), so \( \tan^{-1}(1) = \frac{\pi}{4} \). Step 2: Use the formula for the sum of two inverse tangents.
The formula is \( \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \).
Let's apply this to \( \tan^{-1}(2) + \tan^{-1}(3) \), where \( x = 2 \) and \( y = 3 \):
$$\tan^{-1}(2) + \tan^{-1}(3) = \tan^{-1} \left( \frac{2 + 3}{1 - 2 \times 3} \right) = \tan^{-1} \left( \frac{5}{1 - 6} \right) = \tan^{-1} \left( \frac{5}{-5} \right) = \tan^{-1}(-1)$$ Step 3: Determine the principal value of \( \tan^{-1}(-1) \).
The principal value of \( \tan^{-1}(-1) \) lies in the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), which is \( -\frac{\pi}{4} \). However, since \( \tan^{-1}(2) \) and \( \tan^{-1}(3) \) are both in \( (0, \frac{\pi}{2}) \), their sum must be in \( (0, \pi) \). The formula used has a caveat when \( xy>1 \). In this case, since \( 2 \times 3 = 6>1 \), we need to adjust the result.
If \( x>0 \) and \( y>0 \) and \( xy>1 \), then \( \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \).
So, \( \tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).
Step 4: Add \( \tan^{-1}(1) \) to the result.
$$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi$$