Inverse Trigonometric Functions: Application Problems (1)

Question

Prove that $\tan^{-1}\dfrac{1}{2} + \tan^{-1}\dfrac{1}{3} = \dfrac{\pi}{4}$.

Solution — Step by Step

Final answer: Proved. The sum equals $\boxed{\mathbf{\pi/4}}$.

Why This Works

The addition formula comes from $\tan(A + B) = (\tan A + \tan B)/(1 - \tan A \tan B)$. Take $\tan A = a$, $\tan B = b$, then $A + B = \tan^{-1}((a+b)/(1-ab))$ — provided $A + B$ stays in the principal-value range $(-\pi/2, \pi/2)$, which is guaranteed when $ab < 1$ and both $a, b > 0$ (or both negative).

When $ab > 1$, the sum exceeds $\pi/2$ and we need a $\pi$-correction: the formula gives a value in the wrong range, so we add $\pi$ (if both positive) or subtract $\pi$ (if both negative).

Alternative Method

Geometric proof: consider a triangle inscribed in a unit grid. The angles whose tangents are $1/2$ and $1/3$ together fit into a $1 \times 1$ square’s diagonal, forming $45°$ at the corner. This shows the result without algebra.

Common Mistake