Question

(i) If $\tan A=\frac{5}{6}\mathrm{and}\tan B=\frac{1}{11}$, prove that $A+B=\frac{\mathrm{\pi }}{4}$.
(ii) If $\tan A=\frac{m}{m-1}\mathrm{and}\tan B=\frac{1}{2m-1}$, then prove that $A-B=\frac{\mathrm{\pi }}{4}$.

Answer 1

(i)
$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \tan A=\frac{5}{6}\mathrm{and}\tan B=\frac{1}{11} \\ \mathrm{Therefore},\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \Rightarrow \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ \Rightarrow \tan \left(A+B\right)=\frac{{\displaystyle \frac{5}{6}}+{\displaystyle \frac{1}{11}}}{1-{\displaystyle \frac{5}{6}}\times {\displaystyle \frac{1}{11}}} \\ \Rightarrow \tan \left(A+B\right)=\frac{{\displaystyle \frac{61}{66}}}{{\displaystyle \frac{61}{66}}} \\ \Rightarrow \tan \left(A+B\right)=1 \\ \Rightarrow \tan \left(A+B\right)=\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{Therefore},A+B=\frac{\mathrm{\pi }}{4}. \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ =\frac{{\displaystyle \frac{m}{m-1}}-{\displaystyle \frac{1}{2m-1}}}{1+{\displaystyle \frac{m}{(m-1)(2m-1)}}} \\ =\frac{2m^2-m-m+1}{2m^2-m-2m+1+m} \\ =\frac{2m^2-2m+1}{2m^2-2m+1} \\ =1 \\ \Rightarrow A-B={\tan }^{-1}\left(1\right) \\ \Rightarrow A-B=\frac{\mathrm{\pi }}{4} \end{gathered}$$