The discussion confirms that the sum of the inverse tangent functions, specifically $\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}$, equals $\frac{\pi}{4}$. This conclusion is derived by analyzing the product of complex numbers, specifically $z = (2+i)(3+i)$. The mathematical proof demonstrates the efficiency of using complex numbers in trigonometric identities.
PREREQUISITESMathematicians, students studying trigonometry, and anyone interested in the applications of complex numbers in solving trigonometric equations.
Random Variable said:$$(2+i)(3+i) = 5+5i = \sqrt{50}e^{i \arctan (1)} = \sqrt{50}e^{i \frac{\pi}{4}}$$
$$ (2+i)(3+i) = \sqrt{5} e^{i \arctan (\frac{1}{2})} \sqrt{10} e^{i \arctan (\frac{1}{3})} = \sqrt{50} e^{i [\arctan (\frac{1}{2}) + i \arctan(\frac{1}{3})]}$$
Therefore,
$$ \frac{\pi}{4} = \arctan \left( \frac{1}{2}\right) + \arctan \left(\frac{1}{3} \right)$$