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# Solve the following equations for x: (i) tan−12x + tan−13x = nπ + $\frac{3\mathrm{\pi }}{4}$ (ii) tan−1(x + 1) + tan−1(x − 1) = tan−1$\frac{8}{31}$ (iii) ${\mathrm{tan}}^{-1}\frac{1}{4}+2{\mathrm{tan}}^{-1}\frac{1}{5}+{\mathrm{tan}}^{-1}\frac{1}{6}+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}$ (iv) sin−1x + sin−12x = $\frac{\mathrm{\pi }}{3}$ (v) $3{\mathrm{sin}}^{-1}\frac{2x}{1+{x}^{2}}-4{\mathrm{cos}}^{-1}\frac{1-{x}^{2}}{1+{x}^{2}}+2{\mathrm{tan}}^{-1}\frac{2x}{1-{x}^{2}}=\frac{\mathrm{\pi }}{3}$ (vi) ${\mathrm{cos}}^{-1}x+{\mathrm{sin}}^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6}$ (vii) tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x (viii) tan (cos−1x) = sin$\left({\mathrm{cot}}^{-1}\frac{1}{2}\right)$ (ix) tan−1$\left(\frac{1-x}{1+x}\right)-\frac{1}{2}$tan−1x = 0, where x > 0 (x) cot−1x − cot−1(x + 2) = $\frac{\mathrm{\pi }}{12}$, x > 0 (xi) ${\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)+{\mathrm{cot}}^{-1}\left(\frac{1-{x}^{2}}{2x}\right)=\frac{2\mathrm{\pi }}{3},x>0$ (xii) tan−1(x + 2) + tan−1(x − 2) = tan−1$\left(\frac{8}{79}\right)$, x > 0 (xiii) ${\mathrm{tan}}^{-1}\frac{x}{2}+{\mathrm{tan}}^{-1}\frac{x}{3}=\frac{\mathrm{\pi }}{4},0

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Solution

## (i) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}2x+{\mathrm{tan}}^{-1}3x=n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2x+3x}{1-2x×3x}\right)=n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{5x}{1-6{x}^{2}}=\mathrm{tan}\left(n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{5x}{1-6{x}^{2}}=-1\phantom{\rule{0ex}{0ex}}⇒5x=-1+6{x}^{2}\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}-5x-1=0\phantom{\rule{0ex}{0ex}}⇒\left(6x+1\right)\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}⇒x=-\frac{1}{6}\left[\mathrm{As}x=1\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (ii) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\left(x+1\right)+{\mathrm{tan}}^{-1}\left(x-1\right)={\mathrm{tan}}^{-1}\frac{8}{31}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left\{\frac{x+1+x-1}{1-\left(x+1\right)×\left(x-1\right)}\right\}={\mathrm{tan}}^{-1}\frac{8}{31}\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{1-{x}^{2}+1}=\frac{8}{31}\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{2-{x}^{2}}=\frac{8}{31}\phantom{\rule{0ex}{0ex}}⇒31x=8-4{x}^{2}\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+31x-8=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+32x-x-8=0\phantom{\rule{0ex}{0ex}}⇒\left(4x-1\right)\left(x+8\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{4}\left[\mathrm{As}x=-8\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (iii) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\frac{1}{4}+2{\mathrm{tan}}^{-1}\frac{1}{5}+{\mathrm{tan}}^{-1}\frac{1}{6}+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\frac{1}{4}+{\mathrm{tan}}^{-1}\frac{1}{5}+{\mathrm{tan}}^{-1}\frac{1}{5}+{\mathrm{tan}}^{-1}\frac{1}{6}+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{4}+\frac{1}{5}}{1-\frac{1}{4}×\frac{1}{5}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{6}}{1-\frac{1}{5}×\frac{1}{6}}\right)+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{9}{20}}{\frac{19}{20}}\right)+{\mathrm{tan}}^{-1}\left(\frac{\frac{11}{30}}{\frac{29}{30}}\right)+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{9}{19}\right)+{\mathrm{tan}}^{-1}\left(\frac{11}{29}\right)+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{9}{19}+\frac{11}{29}}{1-\frac{11}{29}×\frac{9}{19}}\right)+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{235}{226}\right)+{\mathrm{tan}}^{-1}\frac{1}{x}=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{235}{226}+\frac{1}{x}}{1-\frac{235}{226}×\frac{1}{x}}\right)=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{235x+226}{226x-235}=\mathrm{tan}\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{235x+226}{226x-235}=1\phantom{\rule{0ex}{0ex}}⇒235x+226=226x-235\phantom{\rule{0ex}{0ex}}⇒9x=-461\phantom{\rule{0ex}{0ex}}⇒x=-\frac{461}{9}$ (iv) We know ${\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}y={\mathrm{sin}}^{-1}\left[x\sqrt{1-{y}^{2}}+y\sqrt{1-{x}^{2}}\right]$ $\therefore {\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}2x=\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}x+{\mathrm{sin}}^{-1}2x={\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}x-{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)=-{\mathrm{sin}}^{-1}2x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}\left[x\sqrt{1-\frac{3}{4}}+\frac{\sqrt{3}}{2}\sqrt{1-{x}^{2}}\right]=-{\mathrm{sin}}^{-1}2x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-{x}^{2}}\right]={\mathrm{sin}}^{-1}\left(-2x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-{x}^{2}}=-2x\phantom{\rule{0ex}{0ex}}⇒x+\sqrt{3}\sqrt{1-{x}^{2}}=-4x\phantom{\rule{0ex}{0ex}}⇒5x=-\sqrt{3}\sqrt{1-{x}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\phantom{\rule{0ex}{0ex}}25{x}^{2}=3-3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒28{x}^{2}=3\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{2}\sqrt{\frac{3}{7}}$ (v) $3{\mathrm{sin}}^{-1}\left(\frac{2x}{1+{x}^{2}}\right)-4{\mathrm{cos}}^{-1}\left(\frac{1-{x}^{2}}{1+{x}^{2}}\right)+2{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)=\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒6{\mathrm{tan}}^{-1}x-8{\mathrm{tan}}^{-1}x+4{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{3}\left[\because 2{\mathrm{tan}}^{-1}x={\mathrm{sin}}^{-1}\left(\frac{2x}{1+{x}^{2}}\right),2{\mathrm{tan}}^{-1}x={\mathrm{cos}}^{-1}\left(\frac{1-{x}^{2}}{1+{x}^{2}}\right)\mathrm{and}2{\mathrm{tan}}^{-1}x={\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)\right]\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{tan}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}$ $\therefore x=\frac{1}{\sqrt{3}}$ (vi) ${\mathrm{cos}}^{-1}x+{\mathrm{sin}}^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x+{\mathrm{sin}}^{-1}\frac{x}{2}={\mathrm{sin}}^{-1}\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x={\mathrm{sin}}^{-1}\frac{1}{2}-{\mathrm{sin}}^{-1}\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x={\mathrm{sin}}^{-1}\left[\frac{1}{2}\sqrt{1-\frac{{x}^{2}}{4}}-\frac{x}{2}\sqrt{1-\frac{1}{4}}\right]\left[\because {\mathrm{sin}}^{-1}x-{\mathrm{sin}}^{-1}y={\mathrm{sin}}^{-1}\left\{x\sqrt{1-y}-y\sqrt{1-{x}^{2}}\right\}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{cos}}^{-1}x={\mathrm{sin}}^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{-1}\sqrt{1-{x}^{2}}={\mathrm{sin}}^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right]\phantom{\rule{0ex}{0ex}}⇒\sqrt{1-{x}^{2}}=0\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\phantom{\rule{0ex}{0ex}}⇒1-{x}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x=±1\left[\mathrm{As}x=-1\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (vii) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ and ${\mathrm{tan}}^{-1}x-{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x-y}{1+xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\left(x+1\right)+{\mathrm{tan}}^{-1}\left(x-1\right)+{\mathrm{tan}}^{-1}x={\mathrm{tan}}^{-1}3x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left\{\frac{x+1+x-1}{1-\left(x+1\right)×\left(x+1\right)}\right\}={\mathrm{tan}}^{-1}3x-{\mathrm{tan}}^{-1}x\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2x}{2-{x}^{2}}\right)={\mathrm{tan}}^{-1}\left(\frac{3x-x}{1+3{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{2-{x}^{2}}=\frac{2x}{1+3{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒2-{x}^{2}=1+3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}-1=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{1}{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (viii) $\mathrm{tan}\left({\mathrm{cos}}^{-1}x\right)=\mathrm{sin}\left({\mathrm{cot}}^{-1}\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left({\mathrm{cos}}^{-1}x\right)=\mathrm{sin}\left({\mathrm{tan}}^{-1}2\right)\left[\because {\mathrm{cot}}^{-1}x={\mathrm{tan}}^{-1}\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\left({\mathrm{tan}}^{-1}\frac{\sqrt{1-{x}^{2}}}{x}\right)=\mathrm{sin}\left({\mathrm{sin}}^{-1}\frac{2}{\sqrt{1+4}}\right)\left[\because {\mathrm{cos}}^{-1}x={\mathrm{tan}}^{-1}\frac{\sqrt{1-{x}^{2}}}{x}\mathrm{and}{\mathrm{tan}}^{-1}x={\mathrm{sin}}^{-1}\frac{x}{\sqrt{1+{x}^{2}}}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{\sqrt{1-{x}^{2}}}{x}=\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}\mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\phantom{\rule{0ex}{0ex}}⇒\frac{1-{x}^{2}}{{x}^{2}}=\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒5-5{x}^{2}=4{x}^{2}\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}=5\phantom{\rule{0ex}{0ex}}⇒x=±\frac{\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (ix) ${\mathrm{tan}}^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2}{\mathrm{tan}}^{-1}\left(x\right)=0\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\mathrm{tan}}^{-1}\left(x\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}1-{\mathrm{tan}}^{-1}x=\frac{1}{2}{\mathrm{tan}}^{-1}\left(x\right)\left[\because {\mathrm{tan}}^{-1}1-{\mathrm{tan}}^{-1}x={\mathrm{tan}}^{-1}\left(\frac{1-x}{1+x}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}1=\frac{3}{2}{\mathrm{tan}}^{-1}\left(x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }}{4}=\frac{3}{2}{\mathrm{tan}}^{-1}\left(x\right)\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }}{6}={\mathrm{tan}}^{-1}\left(x\right)\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (x) $⇒{\mathrm{cot}}^{-1}\left(x\right)-{\mathrm{cot}}^{-1}\left(x+2\right)=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)+{\mathrm{cot}}^{-1}\left(\frac{1}{x+2}\right)=\frac{\mathrm{\pi }}{12}\left[\because {\mathrm{cot}}^{-1}x={\mathrm{tan}}^{-1}\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\frac{1}{x\left(x+2\right)}}\right)=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{2}{x\left(x+2\right)}}{\frac{{x}^{2}+2x+1}{x\left(x+2\right)}}\right)=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2}{{x}^{2}+2x+1}\right)=\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\mathrm{tan}\frac{\mathrm{\pi }}{12}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\mathrm{tan}\left(\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\frac{\mathrm{tan}\frac{\mathrm{\pi }}{3}-\mathrm{tan}\frac{\mathrm{\pi }}{4}}{1+\mathrm{tan}\frac{\mathrm{\pi }}{3}×\mathrm{tan}\frac{\mathrm{\pi }}{4}}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}⇒\left(\frac{2}{{x}^{2}+2x+1}\right)=\frac{2}{{\left(\sqrt{3}+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{\left(x+1\right)}^{2}}=\frac{1}{{\left(\sqrt{3}+1\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒x+1=\sqrt{3}+1\phantom{\rule{0ex}{0ex}}⇒x=\sqrt{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (xi) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)+{\mathrm{cot}}^{-1}\left(\frac{1-{x}^{2}}{2x}\right)=\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)+{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)=\frac{2\mathrm{\pi }}{3}\left[\because {\mathrm{cot}}^{-1}x={\mathrm{tan}}^{-1}\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)=\frac{\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒2{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{3}\left[\because 2{\mathrm{tan}}^{-1}x{\mathrm{tan}}^{-1}\left(\frac{2x}{1-{x}^{2}}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}x=\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\mathrm{tan}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (xii) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\left(x+2\right)+{\mathrm{tan}}^{-1}\left(x-2\right)={\mathrm{tan}}^{-1}\frac{8}{79}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{x+2+x-2}{1-\left(x+2\right)×\left(x-2\right)}\right)={\mathrm{tan}}^{-1}\frac{8}{79}\phantom{\rule{0ex}{0ex}}⇒\frac{2x}{1-{x}^{2}+4}=\frac{8}{79}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{5-{x}^{2}}=\frac{4}{79}\phantom{\rule{0ex}{0ex}}⇒79x=20-4{x}^{2}\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+79x-20=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+80x-x-20=0\phantom{\rule{0ex}{0ex}}⇒\left(4x-1\right)\left(x+20\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{4}\mathrm{or}-20\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore x=\frac{1}{4}\left[\because x>0\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ (xiii) We know ${\mathrm{tan}}^{-1}x+{\mathrm{tan}}^{-1}y={\mathrm{tan}}^{-1}\left(\frac{x+y}{1-xy}\right)$ $\therefore {\mathrm{tan}}^{-1}\left(\frac{x}{2}\right)+{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{x}{2}+\frac{x}{3}}{1-\frac{x}{2}×\frac{x}{3}}\right)=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{tan}}^{-1}\left(\frac{\frac{5x}{6}}{\frac{6-{x}^{2}}{6}}\right)=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{5x}{6-{x}^{2}}=\mathrm{tan}\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{5x}{6-{x}^{2}}=1\phantom{\rule{0ex}{0ex}}⇒5x=6-{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5x-6=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(x+6\right)=0\phantom{\rule{0ex}{0ex}}⇒x=1\left[\because 0

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