If in a right angled triangle the greatest side is $a$, then $\tan\left(\frac{C}{2}\right) =$
Step-by-Step Solution
Key Concept: Use half-angle substitution and the constraint from triangle geometry to determine which form of $\tan(C/2)$ is valid.
Given a right triangle with $\angle A = 90°$, the greatest side is $a$ (the hypotenuse), so $b^2 + c^2 = a^2$. Using the half-angle formula, $\sin C = \frac{c}{a} = \frac{2\tan(C/2)}{1+\tan^2(C/2)}$. Setting $\tan(C/2) = t$ and simplifying the constraint $\frac{a+b}{c} = 1$ (from triangle inequality), we get $2at + c t^2 = c t^2 - 2at + c$, which yields $t = \frac{a+b}{c}$. The condition that $\tan(C/2) < 1$ (equivalently $C < \pi/2$) is impossible when $\frac{a+b}{c} \geq 1$, so we must have $\tan(C/2) = \frac{a-b}{c}$.
Correct Answer: 1