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Question

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

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Solution

## (i) $\mathrm{We}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\mathrm{tan}A=\frac{5}{6}\mathrm{and}\mathrm{tan}B=\frac{1}{11}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{tan}\left(A+B\right)=\frac{\mathrm{tan}A+\mathrm{tan}B}{1-\mathrm{tan}A\mathrm{tan}B}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left(A+B\right)=\frac{\mathrm{tan}A+\mathrm{tan}B}{1-\mathrm{tan}A\mathrm{tan}B}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left(A+B\right)=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6}×\frac{1}{11}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left(A+B\right)=\frac{\frac{61}{66}}{\frac{61}{66}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left(A+B\right)=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left(A+B\right)=\mathrm{tan}\left(\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},A+B=\frac{\mathrm{\pi }}{4}.\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}.\phantom{\rule{0ex}{0ex}}$ (ii) $\mathrm{We}\mathrm{know}\mathrm{that}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(A-B\right)=\frac{\mathrm{tan}A-\mathrm{tan}B}{1+\mathrm{tan}A\mathrm{tan}B}\phantom{\rule{0ex}{0ex}}=\frac{\frac{m}{m-1}-\frac{1}{2m-1}}{1+\frac{m}{\left(m-1\right)\left(2m-1\right)}}\phantom{\rule{0ex}{0ex}}=\frac{2{m}^{2}-m-m+1}{2{m}^{2}-m-2m+1+m}\phantom{\rule{0ex}{0ex}}=\frac{2{m}^{2}-2m+1}{2{m}^{2}-2m+1}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}⇒A-B={\mathrm{tan}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}⇒A-B=\frac{\mathrm{\pi }}{4}$

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