Question

With usual notations in \( \triangle ABC \), if \( C = 90^\circ \), then \( \tan^{-1}\left(\dfrac{a}{b+c}\right) + \tan^{-1}\left(\dfrac{b}{c+a}\right) \) is

• A. \( \dfrac{\pi}{4} \)
• B. \( \dfrac{\pi}{6} \)
• C. \( \pi \)
• D. \( \dfrac{\pi}{3} \)

Hint:

In right-angled triangles, always use \( c^2 = a^2 + b^2 \) to simplify inverse trigonometric expressions.

Answer 1

The Correct Option is A

Step 1: Use right triangle relations.
Since \( C = 90^\circ \), we have \[ c^2 = a^2 + b^2 \] Step 2: Use the identity for sum of inverse tangents.
\[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \quad \text{when } xy<1 \] Step 3: Apply the formula.
Let \[ x = \frac{a}{b+c}, \quad y = \frac{b}{c+a} \] Step 4: Compute numerator and denominator.
\[ \frac{x+y}{1-xy} = \frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \frac{ab}{(b+c)(c+a)}} \] Step 5: Simplify using \( c^2 = a^2 + b^2 \).
After simplification, \[ \tan^{-1}(1) = \frac{\pi}{4} \] Step 6: Conclusion.
Hence, the required value is \( \dfrac{\pi}{4} \).