Question

Solve the following equations for x:

(i) ${\tan }^{-1}\frac{1}{4}+2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{6}+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}$

(ii) $3{\sin }^{-1}\frac{2x}{1+x^2}-4{\cos }^{-1}\frac{1-x^2}{1+x^2}+2{\tan }^{-1}\frac{2x}{1-x^2}=\frac{\mathrm{\pi }}{3}$

(iii) ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\mathrm{\pi }}{3},x>0$

(iv)

Answer 1

(i) We know
${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

$$\begin{gathered} \therefore {\tan }^{-1}\frac{1}{4}+2{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{6}+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{6}+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}\times \frac{1}{5}}\right)+{\tan }^{-1}\left(\frac{\frac{1}{5}+\frac{1}{6}}{1-\frac{1}{5}\times \frac{1}{6}}\right)+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{9}{20}}{\frac{19}{20}}\right)+{\tan }^{-1}\left(\frac{\frac{11}{30}}{\frac{29}{30}}\right)+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{9}{19}\right)+{\tan }^{-1}\left(\frac{11}{29}\right)+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{9}{19}+\frac{11}{29}}{1-\frac{11}{29}\times \frac{9}{19}}\right)+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{235}{226}\right)+{\tan }^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\frac{235}{226}\times \frac{1}{x}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{235x+226}{226x-235}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{235x+226}{226x-235}=1 \\ \Rightarrow 235x+226=226x-235 \\ \Rightarrow 9x=-461 \\ \Rightarrow x=-\frac{461}{9} \end{gathered}$$

(ii)

$$\begin{gathered} 3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\mathrm{\pi }}{3} \\ \Rightarrow 6{\tan }^{-1}x-8{\tan }^{-1}x+4{\tan }^{-1}x=\frac{\mathrm{\pi }}{3}\left[\because 2{\tan }^{-1}x={\sin }^{-1}\left(\frac{2x}{1+x^2}\right),2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\mathrm{and}2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right] \\ \Rightarrow 2{\tan }^{-1}x=\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{6} \\ \Rightarrow x=\tan \frac{\mathrm{\pi }}{6} \\ \Rightarrow x=\frac{1}{\sqrt{3}} \end{gathered}$$

(iii) We know
${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

$$\begin{gathered} \therefore {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\mathrm{\pi }}{3}\left[\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x}\right] \\ \Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\mathrm{\pi }}{3} \\ \Rightarrow 2{\tan }^{-1}x=\frac{\mathrm{\pi }}{3}\left[\because 2{\tan }^{-1}x{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\right] \\ \Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{6} \\ \Rightarrow x=\tan \frac{\mathrm{\pi }}{6} \\ \Rightarrow x=\frac{1}{\sqrt{3}} \end{gathered}$$

(iv)