Solve the following equations for x:

Solution:

(i) We know

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

$\therefore \tan ^{-1} 2 x+\tan ^{-1} 3 x=n \pi+\frac{3 \pi}{4}$

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

$\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \left(n \pi+\frac{3 \pi}{4}\right)$

$\Rightarrow \frac{5 x}{1-6 x^{2}}=-1$

$\Rightarrow 5 x=-1+6 x^{2}$

$\Rightarrow 6 x^{2}-5 x-1=0$

$\Rightarrow(6 x+1)(x-1)=0$

$\Rightarrow x=-\frac{1}{6} \quad[$ As $x=1$ is not satisfying the equation $]$

(ii)  We know

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

$\therefore \tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}$

$\Rightarrow \tan ^{-1}\left\{\frac{x+1+x-1}{1-(x+1) \times(x-1)}\right\}=\tan ^{-1} \frac{8}{31}$

$\Rightarrow \frac{2 x}{1-x^{2}+1}=\frac{8}{31}$

$\Rightarrow \frac{2 x}{2-x^{2}}=\frac{8}{31}$

$\Rightarrow 31 x=8-4 x^{2}$

$\Rightarrow 4 x^{2}+31 x-8=0$

$\Rightarrow 4 x^{2}+32 x-x-8=0$

$\Rightarrow(4 x-1)(x+8)=0$

$\Rightarrow x=\frac{1}{4} \quad[$ As $x=-8$ is not satisfying the equation $]$

(iii) We know

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

$\therefore \tan ^{-1}(x+1)+\tan ^{-1}(x-1)+\tan ^{-1} x=\tan ^{-1} 3 x$

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

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

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

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

$\Rightarrow 4 x^{2}-1=0$

$\Rightarrow x^{2}=\frac{1}{4}$

$\Rightarrow x=\pm \frac{1}{2}$

(iv)

$\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1}(x)=0$

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

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

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

$\Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1}(x)$

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

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

(v)

$\Rightarrow \cot ^{-1}(x)-\cot ^{-1}(x+2)=\frac{\pi}{12}$

$\Rightarrow \tan ^{-1}\left(\frac{1}{x}\right)+\cot ^{-1}\left(\frac{1}{x+2}\right)=\frac{\pi}{12}\left[\because \cot ^{-1} x=\tan ^{-1} \frac{1}{x}\right]$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\frac{1}{x(x+2)}}\right)=\frac{\pi}{12}$

$\Rightarrow \tan ^{-1}\left(\frac{\frac{2}{x(x+2)}}{\frac{x^{2}+2 x+1}{x(x+2)}}\right)=\frac{\pi}{12}$

$\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}+2 x+1}\right)=\frac{\pi}{12}$

$\Rightarrow\left(\frac{2}{x^{2}+2 x+1}\right)=\tan \frac{\pi}{12}$

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

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

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

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

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

$\Rightarrow \frac{1}{(x+1)^{2}}=\frac{1}{(\sqrt{3}+1)^{2}}$

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

$\Rightarrow x=\sqrt{3}$

(vi) We know

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

$\therefore \tan ^{-1}(x+2)+\tan ^{-1}(x-2)=\tan ^{-1} \frac{8}{79}$

$\Rightarrow \tan ^{-1}\left(\frac{x+2+x-2}{1-(x+2) \times(x-2)}\right)=\tan ^{-1} \frac{8}{79}$

$\Rightarrow \frac{2 x}{1-x^{2}+4}=\frac{8}{79}$

$\Rightarrow \frac{x}{5-x^{2}}=\frac{4}{79}$

$\Rightarrow 79 x=20-4 x^{2}$

$\Rightarrow 4 x^{2}+79 x-20=0$

$\Rightarrow 4 x^{2}+80 x-x-20=0$

$\Rightarrow(4 x-1)(x+20)=0$

$\Rightarrow x=\frac{1}{4}$ or $-20$

$\therefore x=\frac{1}{4} \quad[\because x>0]$

(vii)  We know

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

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

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

$\Rightarrow \tan ^{-1}\left(\frac{\frac{\frac{6 x}{6}}{6-x^{2}}}{6}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{5 x}{6-x^{2}}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{5 x}{6-x^{2}}=1$

$\Rightarrow 5 x=6-x^{2}$

$\Rightarrow x^{2}+5 x-6=0$

$\Rightarrow(x-1)(x+6)=0$

$\Rightarrow x=1 \quad[\because 0

(viii)
We know

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

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

$\Rightarrow \tan ^{-1}\left(\frac{\frac{x^{2}+2 x-8+x^{2}-2 x-8}{(x-4)(x+4)}}{\frac{x^{2}-16-x^{2}+4}{(x-4)(x+4)}}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{2 x^{2}-16}{-12}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{2 x^{2}-16}{-12}=1$

$\Rightarrow 2 x^{2}-16=-12$

$\Rightarrow 2 x^{2}=4$

$\Rightarrow x^{2}=2$

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

(ix) 
We know

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

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

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

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

$\Rightarrow-6+2 x^{2}=12$

$\Rightarrow 2 x^{2}=18$

$\Rightarrow x^{2}=9$

$\Rightarrow x=\pm 3$

(x)

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

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

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

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

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

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

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

$\Rightarrow 2 x^{2}=7$

$\Rightarrow x^{2}=\frac{7}{2}$

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