Question

If A and B are acute angles such that tan A = $\frac{1}{3}$, tan B = $\frac{1}{2}$ and tan (A + B) = $\frac{\tan A+\tan B}{1-\tan A\tan B}$ , show that A + B = 45°.

Answer 1

Given:
$$\begin{gathered} \tan A=\frac{1}{3}\mathrm{and}\tan B=\frac{1}{2} \\ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \mathrm{On}\mathrm{substituting}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{RHS}\mathrm{of}\mathrm{the}\mathrm{expression},\mathrm{we}\mathrm{get}: \\ \frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}}\right)}{\left(1-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{2}}\right)}=\frac{\left({\displaystyle \frac{5}{6}}\right)}{\left(1-{\displaystyle \frac{1}{6}}\right)}=\frac{\left({\displaystyle \frac{5}{6}}\right)}{\left({\displaystyle \frac{5}{6}}\right)}=1 \\ \Rightarrow \tan (A+B)=1=\tan {45}^{\mathrm{o}}[\because \tan {45}^{\mathrm{o}}=1] \\ \therefore \mathrm{A}+\mathrm{B}={45}^{\mathrm{o}} \end{gathered}$$

$$\begin{gathered} \tan A=\frac{1}{3}and\tan B=\frac{1}{2} \\ \mathrm{By}\mathrm{substituting}\mathrm{the}\mathrm{values},\mathrm{we}\mathrm{get}; \\ \tan (\mathrm{A}+\mathrm{B})=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A}\tan \mathrm{B}}=\frac{\left({\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{2}}\right)}{\left(1-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{2}}\right)}=\frac{\left({\displaystyle \frac{5}{6}}\right)}{\left(1-{\displaystyle \frac{1}{6}}\right)}=\frac{\left({\displaystyle \frac{5}{6}}\right)}{\left({\displaystyle \frac{5}{6}}\right)}=1 \\ \tan (\mathrm{A}+\mathrm{B})=1=\tan {45}^{\mathrm{o}} \\ \mathrm{Hence},\mathrm{A}+\mathrm{B}=45 \end{gathered}$$13