(i) If tan A=


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

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


(i) We have:

$\tan A=\frac{5}{6}$ and $\tan B=\frac{1}{11}$

Therefore, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A+B)=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6} \times \frac{1}{11}}$

$\Rightarrow \tan (A+B)=\frac{\frac{61}{66}}{\frac{61}{66}}$

$\Rightarrow \tan (A+B)=1$

$\Rightarrow \tan (A+B)=\tan \left(\frac{\pi}{4}\right)$

Therefore, $A+B=\frac{\pi}{4}$.

Hence proved.

(ii) We know that

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$

$=\frac{\frac{m}{m-1}-\frac{1}{2 m-1}}{1+\frac{m}{(m-1)(2 m-1)}}$

$=\frac{2 m^{2}-m-m+1}{2 m^{2}-m-2 m+1+m}$

$=\frac{2 m^{2}-2 m+1}{2 m^{2}-2 m+1}$


$\Rightarrow A-B=\tan ^{-1}(1)$

$\Rightarrow A-B=\frac{\pi}{4}$

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