# What is the value of $\tan^{-1}\large \frac{2}{11}$$+tan^{-1} \large \frac{7}{24}? ## 1 Answer Need homework help? Click here. Toolbox: • $$tan^{-1}x+tan^{-1}y=tan^{-1}\Large \frac{x+y}{1-xy}$$ $$\:if\:xy < 1$$ Given \tan^{-1}\large \frac{2}{11}$$ +tan^{-1} \large \frac{7}{24}$
Let $x = \large \frac{2}{11}$ and $y = \large \frac{7}{24}$.
We know that $$tan^{-1}x+tan^{-1}y=tan^{-1}\Large \frac{x+y}{1-xy}$$ $$\:if\:xy < 1$$
$\Rightarrow x+y = \large \frac{2}{11} + \frac{7}{24}$$= \large \frac{24 \times 2}{24 \times 11} + \frac{11 \times 7}{11 \times 24}$$ $$= \large \frac{48+77}{264} = \large\frac{125}{264} \Rightarrow 1 - xy = 1 -$$\large \frac{2}{11} \times \frac{7}{24}$$= 1 -$$\large\frac{14}{264} = \frac{264 - 14}{264} = \frac{250}{264}$
$\Rightarrow \large\frac{x+y}{1-xy}$$=$$\large\frac{125}{264}$$\times$${\large\frac{264}{250}}$$=$$\large\frac{125}{250}$ $=$$\large \frac{1}{2} Substituting \large\frac{x+y}{1-xy} = \frac{1}{2}, we get \tan^{-1}\large \frac{2}{11} +$$tan^{-1} \large \frac{7}{24}$ = $tan^{-1}\large \frac{1}{2}$