Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$

Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.

Attempt:

I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.

Solution 1:

A proof (almost) without words:

The sum between the red angle and the blue angle is the complementary angle of $\frac{\pi}{4}$, since in the figure we have a right isosceles triangle.

Solution 2:

Sine addition identity: $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$

Cosine addition identity: $$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta.$$

From the above, we obtain the tangent addition identity: $$\begin{align*} \tan (\alpha + \beta) &= \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \\ &= \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} \\ &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}. \end{align*}$$

From this, we now let $x = \tan \alpha$, $y = \tan \beta$, or equivalently, $\alpha = \tan^{-1} x$, $\beta = \tan^{-1} y$, and substitute: $$\tan(\tan^{-1} x + \tan^{-1} y) = \frac{x+y}{1-xy}.$$ Now taking the inverse tangent of both sides, we obtain the inverse tangent identity: $$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}.$$ Now let $x = 1/2$, $y = 1/3$, and simplify the right hand side.

Solution 3:

$$\tan[\tan^{-1}(1/2)+\tan^{-1}(1/3)]=\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}$$

Can you continue?