Question

(i) $2{\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{24}{7}$
(ii) ${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\cos }^{-1}\frac{3}{5}=\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$
(iii) ${\tan }^{-1}\frac{2}{3}=\frac{1}{2}{\tan }^{-1}\frac{12}{5}$

(iv) ${\tan }^{-1}\frac{1}{7}+2{\tan }^{-1}\frac{1}{3}=\frac{\mathrm{\pi }}{4}$

(v) ${\sin }^{-1}\frac{4}{5}+2{\tan }^{-1}\frac{1}{3}=\frac{\mathrm{\pi }}{2}$
(vi) $2{\sin }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{17}{31}=\frac{\mathrm{\pi }}{4}$
(vii) $2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{4}{7}$
(viii) $2{\tan }^{-1}\frac{3}{4}-{\tan }^{-1}\frac{17}{31}=\frac{\mathrm{\pi }}{4}$
(ix) $2{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)={\tan }^{-1}\left(\frac{31}{17}\right)$
(x) $4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{239}=\frac{\mathrm{\pi }}{4}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{LHS}=2{\sin }^{-1}\frac{3}{5} \\ =2{\tan }^{-1}\frac{{\displaystyle \frac{3}{4}}}{\sqrt{1-{\displaystyle \frac{9}{25}}}}\left[\because {\sin }^{-1}x={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\right] \\ =2{\tan }^{-1}\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}} \\ =2{\tan }^{-1}\frac{3}{4} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{3}{4}}}{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}\right\}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{7}{16}}}\right\} \\ ={\tan }^{-1}=\mathrm{RHS} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{LHS}={\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{2}{9}}}{1-{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{9}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{17}{36}}}{{\displaystyle \frac{34}{36}}}\right) \\ ={\tan }^{-1}\frac{1}{2} \\ =\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\displaystyle \frac{1}{4}}}{1+{\displaystyle \frac{1}{4}}}\right)\left[\because {\tan }^{-1}x=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \\ =\frac{1}{2}{\cos }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{5}{4}}}\right) \\ =\frac{1}{2}{\cos }^{-1}\left(\frac{{\displaystyle 3}}{5}\right) \\ \mathrm{Now}, \\ {\tan }^{-1}\frac{1}{2}=\frac{1}{2}{\sin }^{-1}\left(\frac{{\displaystyle \frac{2}{2}}}{1+{\displaystyle \frac{1}{4}}}\right)\left[\because {\tan }^{-1}\mathrm{x}=\frac{1}{2}{\sin }^{-1}\left(\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\right)\right] \\ =\frac{1}{2}{\sin }^{-1}\left(\frac{1}{{\displaystyle \frac{5}{4}}}\right) \\ =\frac{1}{2}{\sin }^{-1}\left(\frac{4}{{\displaystyle 5}}\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{LHS}={\tan }^{-1}\frac{2}{3} \\ =\frac{1}{2}{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{2}{3}}}{1-{\left({\displaystyle \frac{2}{3}}\right)}^2}\right\}\left[\because {\tan }^{-1}x=\frac{1}{2}{\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ =\frac{1}{2}{\tan }^{-1}\left\{\frac{{\displaystyle \frac{4}{3}}}{{\displaystyle \frac{5}{9}}}\right\} \\ =\frac{1}{2}{\tan }^{-1}\frac{12}{5}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{LHS}={\tan }^{-1}\frac{1}{7}+2{\tan }^{-1}\frac{1}{3} \\ ={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\right\}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\left\{\frac{{\displaystyle \frac{{\displaystyle 2}}{3}}}{{\displaystyle \frac{8}{9}}}\right\} \\ ={\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{3}{4} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{1}{7}}\times {\displaystyle \frac{3}{4}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{25}{28}}}{{\displaystyle \frac{25}{28}}}\right) \\ ={\tan }^{-1}1=\frac{\mathrm{\pi }}{4}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{LHS}={\sin }^{-1}\frac{4}{5}+2{\tan }^{-1}\frac{1}{3} \\ ={\sin }^{-1}\frac{4}{5}+{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\left({\displaystyle \frac{1}{3}}\right)}^2}\right\}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\sin }^{-1}\frac{4}{5}+{\tan }^{-1}\left\{\frac{{\displaystyle \frac{{\displaystyle 2}}{3}}}{{\displaystyle \frac{8}{9}}}\right\} \\ ={\sin }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{3}{4} \\ ={\sin }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{1}{\sqrt{1+{\displaystyle \frac{9}{16}}}}\left[\because {\tan }^{-1}x={\cos }^{-1}\frac{1}{\sqrt{1+{\displaystyle x^2}}}\right] \\ ={\sin }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{1}{{\displaystyle \frac{5}{4}}} \\ ={\sin }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{4}{5} \\ =\frac{\mathrm{\pi }}{2}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{LHS}=2{\sin }^{-1}\frac{3}{5}-{\tan }^{-1}\frac{17}{31} \\ =2{\tan }^{-1}\frac{{\displaystyle \frac{3}{4}}}{\sqrt{1-{\displaystyle \frac{9}{25}}}}-{\tan }^{-1}\frac{17}{31}\left[\because {\sin }^{-1}x={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}\right] \\ =2{\tan }^{-1}\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}-{\tan }^{-1}\frac{17}{31} \\ =2{\tan }^{-1}\frac{3}{4}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{3}{4}}}{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}\right\}-{\tan }^{-1}\frac{17}{31}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{7}{16}}}\right\}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\frac{24}{7}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{24}{7}}-{\displaystyle \frac{17}{31}}}{1+{\displaystyle \frac{24}{7}}\times {\displaystyle \frac{17}{31}}}\right)\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}-y}{1+{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{625}{217}}}{{\displaystyle \frac{625}{217}}}\right) \\ ={\tan }^{-1}1=\frac{\mathrm{\pi }}{4}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\mathrm{LHS}=2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{5}}}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2}\right\}+{\tan }^{-1}\frac{1}{8}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle \frac{{\displaystyle 2}}{5}}}{{\displaystyle \frac{24}{25}}}\right\}+{\tan }^{-1}\frac{1}{8} \\ ={\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{1}{8} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{5}{12}}+{\displaystyle \frac{1}{8}}}{1-{\displaystyle \frac{5}{12}}\times {\displaystyle \frac{1}{8}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{13}{24}}}{{\displaystyle \frac{91}{96}}}\right) \\ ={\tan }^{-1}\frac{4}{7}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\mathrm{LHS}=2{\tan }^{-1}\frac{3}{4}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{3}{4}}}{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}\right\}-{\tan }^{-1}\frac{17}{31}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{7}{16}}}\right\}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\frac{24}{7}-{\tan }^{-1}\frac{17}{31} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{24}{7}}-{\displaystyle \frac{17}{31}}}{1+{\displaystyle \frac{24}{7}}\times {\displaystyle \frac{17}{31}}}\right)\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}-y}{1+{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{625}{217}}}{{\displaystyle \frac{625}{217}}}\right) \\ ={\tan }^{-1}1=\frac{\mathrm{\pi }}{4}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\mathrm{LHS}=2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{2}}}{1-{\left({\displaystyle \frac{1}{2}}\right)}^2}\right\}+{\tan }^{-1}\frac{1}{7}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle 1}}{{\displaystyle \frac{3}{4}}}\right\}+{\tan }^{-1}\frac{1}{7} \\ ={\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{7} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{4}{3}}+{\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{4}{3}}\times {\displaystyle \frac{1}{7}}}\right)\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}+y}{1-{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{31}{21}}}{{\displaystyle \frac{17}{21}}}\right) \\ ={\tan }^{-1}\frac{31}{17}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(x\right)\mathrm{LHS}=4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{239} \\ =2{\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{1}{5}}}{1-{\left({\displaystyle \frac{1}{5}}\right)}^2}\right\}-{\tan }^{-1}\frac{1}{239}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ =2{\tan }^{-1}\left\{\frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{24}{25}}}\right\}-{\tan }^{-1}\frac{1}{239} \\ =2{\tan }^{-1}\frac{5}{12}-{\tan }^{-1}\frac{1}{239} \\ ={\tan }^{-1}\left\{\frac{2\times {\displaystyle \frac{5}{12}}}{1-{\left({\displaystyle \frac{5}{12}}\right)}^2}\right\}-{\tan }^{-1}\frac{1}{239}\left[\because 2{\tan }^{-1}x={\tan }^{-1}\left\{\frac{2x}{1-x^2}\right\}\right] \\ ={\tan }^{-1}\left\{\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{119}{144}}}\right\}-{\tan }^{-1}\frac{1}{239} \\ ={\tan }^{-1}\frac{120}{119}-{\tan }^{-1}\frac{1}{239} \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{120}{119}}-{\displaystyle \frac{17}{239}}}{1+{\displaystyle \frac{120}{119}}\times {\displaystyle \frac{1}{239}}}\right)\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{{\displaystyle x}-y}{1+{\displaystyle xy}}\right)\right] \\ ={\tan }^{-1}1=\frac{\mathrm{\pi }}{4}=\mathrm{RHS} \end{gathered}$$