Question

Prove the following results:

(i) ${\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13}={\tan }^{-1}\frac{2}{9}$
(ii) ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\mathrm{\pi }$
(iii) ${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}={\sin }^{-1}\frac{1}{\sqrt{5}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{13} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{1}{13}}}{1-{\displaystyle \frac{1}{7}}\times {\displaystyle \frac{1}{13}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{20}{91}}}{{\displaystyle \frac{90}{91}}}\right) \\ ={\tan }^{-1}\frac{2}{9}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}={\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16} \\ ={\tan }^{-1}\frac{{\displaystyle \frac{12}{13}}}{\sqrt{1-{\displaystyle \frac{144}{169}}}}+{\tan }^{-1}\frac{\sqrt{1-{\displaystyle \frac{16}{25}}}}{{\displaystyle \frac{4}{5}}}+{\tan }^{-1}\frac{63}{16}\left[\because {\sin }^{-1}x={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\mathrm{and}{\cos }^{-1}x={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right] \\ ={\tan }^{-1}\frac{{\displaystyle \frac{12}{13}}}{{\displaystyle \frac{5}{13}}}+{\tan }^{-1}\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}+{\tan }^{-1}\frac{63}{16} \\ ={\tan }^{-1}\frac{{\displaystyle 12}}{5}+{\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{63}{16} \\ =\mathrm{\pi }+{\tan }^{-1}\left(\frac{{\displaystyle \frac{12}{5}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{12}{5}}\times {\displaystyle \frac{3}{4}}}\right)+{\tan }^{-1}\frac{63}{16}\left[\because {\tan }^{-1}x+{\tan }^{-1}y=\mathrm{\pi }+{\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ =\mathrm{\pi }+{\tan }^{-1}\left(\frac{{\displaystyle \frac{63}{20}}}{{\displaystyle \frac{-16}{20}}}\right)+{\tan }^{-1}\frac{63}{16} \\ =\mathrm{\pi }+{\tan }^{-1}\frac{-63}{16}+{\tan }^{-1}\frac{63}{16} \\ =\mathrm{\pi }-{\tan }^{-1}\frac{63}{16}+{\tan }^{-1}\frac{63}{16} \\ =\mathrm{\pi }=\mathrm{RHS} \end{gathered}$$

(iii)
$$\begin{gathered} \mathrm{LHS}={\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{9}}}{1-{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{9}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{17}{36}}}{{\displaystyle \frac{34}{36}}}\right) \\ ={\tan }^{-1}\frac{1}{2} \\ ={\sin }^{-1}\frac{{\displaystyle \frac{1}{2}}}{\sqrt{1+{\left({\displaystyle \frac{1}{2}}\right)}^2}} \\ ={\sin }^{-1}\frac{{\displaystyle 1}}{\sqrt{5}}=\mathrm{RHS} \end{gathered}$$