Solve the following equations for x:

Solution:

(i) We know

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

$\therefore \tan ^{-1} \frac{1}{4}+2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{6}+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{6}+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4} \times \frac{1}{5}}\right)+\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{6}}{1-\frac{1}{5} \times \frac{1}{6}}\right)+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{9}{20}}{\frac{19}{20}}\right)+\tan ^{-1}\left(\frac{\frac{11}{30}}{\frac{29}{30}}\right)+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{9}{19}\right)+\tan ^{-1}\left(\frac{11}{29}\right)+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{9}{19}+\frac{11}{29}}{1-\frac{11}{29} \times \frac{9}{19}}\right)+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{235}{226}\right)+\tan ^{-1} \frac{1}{x}=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\frac{235}{226} \times \frac{1}{x}}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{235 x+226}{226 x-235}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{235 x+226}{226 x-235}=1$

$\Rightarrow 235 x+226=226 x-235$

$\Rightarrow 9 x=-461$

$\Rightarrow x=-\frac{461}{9}$

(ii)

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

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

$\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{3}$

$\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\tan \frac{\pi}{6}$

$\Rightarrow x=\frac{1}{\sqrt{3}}$

(iii) We know

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

$\therefore \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{2 \pi}{3}$

$\Rightarrow \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \quad\left[\because \cot ^{-1} x=\tan ^{-1} \frac{1}{x}\right]$

$\Rightarrow \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$

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

$\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\tan \frac{\pi}{6}$

$\Rightarrow x=\frac{1}{\sqrt{3}}$

(iv) $2 \tan ^{-1}(\sin x)=\tan ^{-1}(2 \sin x), x \neq \frac{\pi}{2}$

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

$\Rightarrow \frac{2 \sin x}{1-\sin ^{2} x}=2 \sin x$

$\Rightarrow 2 \sin x=2 \sin x-2 \sin ^{3} x$

$\Rightarrow 2 \sin ^{3} x=0$

$\Rightarrow \sin x=0$

$\Rightarrow x=0$

(v) $\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}$

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

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

$\Rightarrow 3 \tan ^{-1} x=\frac{2 \pi}{3}$

$\Rightarrow \tan ^{-1} x=\frac{2 \pi}{9}$

$\Rightarrow x=\tan \left(\frac{2 \pi}{9}\right)$

(vi)

$\tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}$

$\Rightarrow \tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\tan ^{-1} 1$

$\Rightarrow \tan ^{-1}\left(\frac{x-2}{x-1}\right)=\tan ^{-1} 1-\tan ^{-1}\left(\frac{x+2}{x+1}\right)$

$\Rightarrow \tan ^{-1}\left(\frac{x-2}{x-1}\right)=\tan ^{-1}\left(\frac{1-\frac{x+2}{x+1}}{1+\frac{x+2}{x+1}}\right)$

$\Rightarrow \tan ^{-1}\left(\frac{x-2}{x-1}\right)=\tan ^{-1}\left(\frac{x+1-x-2}{x+1+x+2}\right)$

$\Rightarrow \tan ^{-1}\left(\frac{x-2}{x-1}\right)=\tan ^{-1}\left(\frac{-1}{2 x+3}\right)$

$\Rightarrow \frac{x-2}{x-1}=\frac{-1}{2 x+3}$

$\Rightarrow 2 x^{2}+3 x-4 x-6=-x+1$

$\Rightarrow 2 x^{2}=1+6$

$\Rightarrow x^{2}=7$

$\Rightarrow x=\pm \sqrt{\frac{7}{2}}$