Question

Prove that \[ 2 \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \frac{\pi}{4} \]

Hint:

Use the addition formula for inverse tangents to simplify expressions involving the sum of multiple inverse tangents.

Answer 1

Step 1: Use the addition formula for inverse tangent.
We use the following identity for the sum of inverse tangents: \[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \] First, let: \[ A = 2 \tan^{-1} \left( \frac{1}{3} \right), \quad B = \tan^{-1} \left( \frac{1}{7} \right) \] Thus, we need to prove: \[ A + B = \frac{\pi}{4} \] Step 2: Applying the addition formula.
Now, apply the addition formula: \[ \tan(A + B) = \frac{\frac{1}{3} + \frac{1}{7}}{1 - \left( \frac{1}{3} \times \frac{1}{7} \right)} \] Simplify the numerator and denominator: \[ \tan(A + B) = \frac{\frac{7 + 3}{21}}{1 - \frac{1}{21}} = \frac{\frac{10}{21}}{\frac{20}{21}} = \frac{10}{20} = \frac{1}{2} \] Step 3: Conclusion.
Since \( \tan \left( \frac{\pi}{4} \right) = 1 \), we conclude that: \[ 2 \tan^{-1} \left( \frac{1}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \frac{\pi}{4} \]