Question

Simplify the inverse trigonometric expression to find its principal value: \[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \] 

• A. \( \frac{\pi}{2} \)
• B. \( \tan^{-1}\left(\frac{2}{5}\right) \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{4} \)

Hint:

Always check the product \( xy \) before using the tangent addition formula. If the product evaluates to greater than one (\( xy > 1 \)), you must use the alternative identity track: \( \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).

Answer 1

The Correct Option is D

Concept: To combine two inverse tangent terms, apply the standard tangent summation identity: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left( \frac{x + y}{1 - xy} \right) \] This identity is valid provided that the product of the arguments satisfies the domain constraint \( xy < 1 \).

Step 1: Verify the domain constraint rule for the arguments.
Identify the arguments from the problem: \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \). Check their product: \[ xy = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \] Since \( \frac{1}{6} < 1 \), the standard summation identity can be applied safely.

Step 2: Substitute the arguments into the summation formula.
Plug the fractions into the identity template: \[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \left(\frac{1}{2} \cdot \frac{1}{3}\right)} \right) \] Simplify the fractional expressions in both the numerator and the denominator: \[ = \tan^{-1}\left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right) = \tan^{-1}\left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) \] This simplifies to a single unit argument: \[ = \tan^{-1}(1) \]

Step 3: Determine the final principal value angle.
The unique angle in the principal branch interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) whose tangent value equals 1 is \( \frac{\pi}{4} \) (\( 40^\circ \)): \[ \tan^{-1}(1) = \frac{\pi}{4} \]