# In any Triangle ABC, if $A=\tan^{-1}2\;and\;B=\tan^{-1}3,$prove that $C=\frac{\pi}{4}$

Toolbox:
• In $$\Delta\: ABC \: A+B+C = \pi$$
• $$tan(A+B)=\large\frac{tanA+tanB}{1-tanA.tanB}$$
• $$tan(\pi-C)=-tanC$$
• $$tan\large\frac{\pi}{4}=1$$
Given ABC is a triangle and $$tanA = 2 \: and\: tanB=3$$
$$\Rightarrow A+B+C=\pi$$
$$\Rightarrow A+B=\pi-C$$

Taking tan on both the sides and using the formula of $$tan(\pi-C)$$
$$\Rightarrow\:tan(A+B)=tan(\pi-C)=-tanC$$
From the above formula of tan(A+B)
$$\Rightarrow\:\large\frac{tanA+tanB}{1-tanAtanB}=-tanC$$

Substituting the values of tanA and tanB we get
$$\large \frac{2+3}{1-2.3}=\large\frac{5}{-5}=-1=-tanC$$
$$\Rightarrow tanC=1$$
$$C=\large\frac{\pi}{4}$$

edited Mar 19, 2013