# Show that

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

(v) $\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$

(vi) $2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

(vii) $2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}$

(viii) $2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$

(ix) $2 \tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\tan ^{-1}\left(\frac{31}{17}\right)$

(x) $4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}$

Solution:

(i) $\mathrm{LHS}=2 \sin ^{-1} \frac{3}{5}$

$=2 \tan ^{-1} \frac{\frac{3}{4}}{\sqrt{1-\frac{9}{25}}} \quad\left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right]$

$=2 \tan ^{-1} \frac{\frac{3}{5}}{\frac{4}{5}}$

$=2 \tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right\}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{\frac{3}{2}}{\frac{2}{10}}\right\}$

$=\tan ^{-1}=$ RHS

(ii) LHS $=\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$

$=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{17}{30}}{\frac{34}{30}}\right)$

$=\tan ^{-1} \frac{1}{2}$

$=\frac{1}{2} \cos ^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right)$                     $\left[\because \tan ^{-1} x=\frac{1}{2} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$

$=\frac{1}{2} \cos ^{-1}\left(\frac{\frac{3}{4}}{\frac{4}{4}}\right)$

$=\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$

Now,

$\tan ^{-1} \frac{1}{2}=\frac{1}{2} \sin ^{-1}\left(\frac{\frac{2}{2}}{1+\frac{1}{4}}\right) \quad\left[\because \tan ^{-1} \mathrm{x}=\frac{1}{2} \sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)\right]$

$=\frac{1}{2} \sin ^{-1}\left(\frac{1}{\frac{5}{4}}\right)$

$=\frac{1}{2} \sin ^{-1}\left(\frac{4}{5}\right)$

(iii) $\mathrm{LHS}=\tan ^{-1} \frac{2}{3}$

$=\frac{1}{2} \tan ^{-1}\left\{\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right\}\left[\because \tan ^{-1} x=\frac{1}{2} \tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\frac{1}{2} \tan ^{-1}\left\{\frac{\frac{4}{3}}{\frac{5}{9}}\right\}$

$=\frac{1}{2} \tan ^{-1} \frac{12}{5}=$ RHS

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

$=\tan ^{-1} \frac{1}{7}+\tan ^{-1}\left\{\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right\}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1} \frac{1}{7}+\tan ^{-1}\left\{\frac{\frac{2}{3}}{\frac{8}{9}}\right\}$

$=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7} \times \frac{3}{4}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{2 \pi}{28}}{\frac{2 \pi}{28}}\right)$

$=\tan ^{-1} 1=\frac{\pi}{4}=$ RHS

(v) LHS $=\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}$

$=\sin ^{-1} \frac{4}{5}+\tan ^{-1}\left\{\frac{2 \times \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right\}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\sin ^{-1} \frac{4}{5}+\tan ^{-1}\left\{\frac{\frac{2}{3}}{\frac{3}{9}}\right\}$

$=\sin ^{-1} \frac{4}{5}+\tan ^{-1} \frac{3}{4}$

$=\sin ^{-1} \frac{4}{5}+\cos ^{-1} \frac{1}{\sqrt{1+\frac{9}{10}}} \quad\left[\because \tan ^{-1} x=\cos ^{-1} \frac{1}{\sqrt{1+x^{2}}}\right]$

$=\sin ^{-1} \frac{4}{5}+\cos ^{-1} \frac{1}{\frac{5}{4}}$

$=\sin ^{-1} \frac{4}{5}+\cos ^{-1} \frac{4}{5}$

$=\frac{\pi}{2}=\mathrm{RHS}$

(vi) LHS $=2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}$

$=2 \tan ^{-1} \frac{\frac{3}{4}}{\sqrt{1-\frac{9}{25}}}-\tan ^{-1} \frac{17}{31} \quad\left[\because \sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^{2}}}\right]$

$=2 \tan ^{-1} \frac{\frac{3}{5}}{\frac{5}{5}}-\tan ^{-1} \frac{17}{31}$

$=2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right\}-\tan ^{-1} \frac{17}{31}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{\frac{3}{2}}{\frac{7}{16}}\right\}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1} \frac{24}{7}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right) \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{025}{217}}{\frac{625}{217}}\right)$

$=\tan ^{-1} 1=\frac{\pi}{4}=\mathrm{RHS}$

(vii) LHS $=2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}+\tan ^{-1} \frac{1}{8}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{\frac{2}{5}}{\frac{24}{25}}\right\}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{95}}\right)$

$=\tan ^{-1} \frac{4}{7}=\mathrm{RHS}$

(viii) LHS $=2 \tan ^{-1} \frac{3}{4}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right\}-\tan ^{-1} \frac{17}{31}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{\frac{3}{2}}{\frac{2}{10}}\right\}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1} \frac{24}{7}-\tan ^{-1} \frac{17}{31}$

$=\tan ^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7} \times \frac{17}{31}}\right) \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{625}{217}}{\frac{625}{217}}\right)$

$=\tan ^{-1} 1=\frac{\pi}{4}=$ RHS

(ix) LHS $=2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right\}+\tan ^{-1} \frac{1}{7}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{1}{\frac{3}{3}}\right\}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

$=\tan ^{-1}\left(\frac{\frac{31}{\frac{21}{12}}}{\frac{17}{21}}\right)$

$=\tan ^{-1} \frac{31}{17}=\mathrm{RHS}$

$(x) \mathrm{LHS}=4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1}\left\{\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right\}-\tan ^{-1} \frac{1}{239}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=2 \tan ^{-1}\left\{\frac{\frac{2}{5}}{\frac{24}{25}}\right\}-\tan ^{-1} \frac{1}{239}$

$=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$

$=\tan ^{-1}\left\{\frac{2 \times \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}\right\}-\tan ^{-1} \frac{1}{239}\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left\{\frac{2 x}{1-x^{2}}\right\}\right]$

$=\tan ^{-1}\left\{\frac{\frac{5}{6}}{\frac{199}{146}}\right\}-\tan ^{-1} \frac{1}{239}$

$=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}$

$=\tan ^{-1}\left(\frac{\frac{120}{199}-\frac{17}{219}}{1+\frac{120}{19} \times \frac{1}{239}}\right) \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\tan ^{-1} 1=\frac{\pi}{4}=$ RHS