Question

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

(iv) sin⁻¹x + sin⁻¹2x = $\frac{\mathrm{\pi }}{3}$

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

(vi) ${\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin$\left({\cot }^{-1}\frac{1}{2}\right)$

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

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹$\left(\frac{8}{79}\right)$, x > 0

(xiii) ${\tan }^{-1}\frac{x}{2}+{\tan }^{-1}\frac{x}{3}=\frac{\mathrm{\pi }}{4},0<x<\sqrt{6}$

Answer 1

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

$$\begin{gathered} \therefore {\tan }^{-1}2x+{\tan }^{-1}3x=n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{2x+3x}{1-2x\times 3x}\right)=n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4} \\ \Rightarrow \frac{5x}{1-6x^2}=\tan \left(n\mathrm{\pi }+\frac{3\mathrm{\pi }}{4}\right) \\ \Rightarrow \frac{5x}{1-6x^2}=-1 \\ \Rightarrow 5x=-1+6x^2 \\ \Rightarrow 6x^2-5x-1=0 \\ \Rightarrow \left(6x+1\right)\left(x-1\right)=0 \\ \Rightarrow x=-\frac{1}{6}\left[\mathrm{As}x=1\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right] \end{gathered}$$


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

$$\begin{gathered} \therefore {\tan }^{-1}\left(x+1\right)+{\tan }^{-1}\left(x-1\right)={\tan }^{-1}\frac{8}{31} \\ \Rightarrow {\tan }^{-1}\left\{\frac{x+1+x-1}{1-\left(x+1\right)\times \left(x-1\right)}\right\}={\tan }^{-1}\frac{8}{31} \\ \Rightarrow \frac{2x}{1-x^2+1}=\frac{8}{31} \\ \Rightarrow \frac{2x}{2-x^2}=\frac{8}{31} \\ \Rightarrow 31x=8-4x^2 \\ \Rightarrow 4x^2+31x-8=0 \\ \Rightarrow 4x^2+32x-x-8=0 \\ \Rightarrow \left(4x-1\right)\left(x+8\right)=0 \\ \Rightarrow x=\frac{1}{4}\left[\mathrm{As}x=-8\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right] \end{gathered}$$


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

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


(iv) We know
${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right]$

$$\begin{gathered} \therefore {\sin }^{-1}x+{\sin }^{-1}2x=\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\sin }^{-1}x+{\sin }^{-1}2x={\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right) \\ \Rightarrow {\sin }^{-1}x-{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=-{\sin }^{-1}2x \\ \Rightarrow {\sin }^{-1}\left[x\sqrt{1-\frac{3}{4}}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right]=-{\sin }^{-1}2x \\ \Rightarrow {\sin }^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\right]={\sin }^{-1}\left(-2x\right) \\ \Rightarrow \frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}=-2x \\ \Rightarrow x+\sqrt{3}\sqrt{1-x^2}=-4x \\ \Rightarrow 5x=-\sqrt{3}\sqrt{1-x^2} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ 25x^2=3-3x^2 \\ \Rightarrow 28x^2=3 \\ \Rightarrow x=\pm \frac{1}{2}\sqrt{\frac{3}{7}} \end{gathered}$$


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

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

(vi)

$$\begin{gathered} {\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\cos }^{-1}x+{\sin }^{-1}\frac{x}{2}={\sin }^{-1}\frac{1}{2} \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\frac{1}{2}-{\sin }^{-1}\frac{x}{2} \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\left[\frac{1}{2}\sqrt{1-\frac{x^2}{4}}-\frac{x}{2}\sqrt{1-\frac{1}{4}}\right]\left[\because {\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}\left\{x\sqrt{1-y}-y\sqrt{1-x^2}\right\}\right] \\ \Rightarrow {\cos }^{-1}x={\sin }^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right] \\ \Rightarrow {\sin }^{-1}\sqrt{1-x^2}={\sin }^{-1}\left[\frac{\sqrt{3}x}{4}-\frac{\sqrt{3}x}{4}\right] \\ \Rightarrow \sqrt{1-x^2}=0 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ \Rightarrow 1-x^2=0 \\ \Rightarrow x=\pm 1\left[\mathrm{As}x=-1\mathrm{is}\mathrm{not}\mathrm{satisfying}\mathrm{the}\mathrm{equation}\right] \end{gathered}$$

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

$$\begin{gathered} \therefore {\tan }^{-1}\left(x+1\right)+{\tan }^{-1}\left(x-1\right)+{\tan }^{-1}x={\tan }^{-1}3x \\ \Rightarrow {\tan }^{-1}\left\{\frac{x+1+x-1}{1-\left(x+1\right)\times \left(x+1\right)}\right\}={\tan }^{-1}3x-{\tan }^{-1}x \\ \Rightarrow {\tan }^{-1}\left(\frac{2x}{2-x^2}\right)={\tan }^{-1}\left(\frac{3x-x}{1+3x^2}\right) \\ \Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2} \\ \Rightarrow 2-x^2=1+3x^2 \\ \Rightarrow 4x^2-1=0 \\ \Rightarrow x^2=\frac{1}{4} \\ \Rightarrow x=\pm \frac{1}{2} \end{gathered}$$


(viii)
$$\begin{gathered} \tan \left({\cos }^{-1}x\right)=\sin \left({\cot }^{-1}\frac{1}{2}\right) \\ \Rightarrow \tan \left({\cos }^{-1}x\right)=\sin \left({\tan }^{-1}2\right)\left[\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x}\right] \\ \Rightarrow \tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)=\sin \left({\sin }^{-1}\frac{2}{\sqrt{1+4}}\right)\left[\because {\cos }^{-1}x={\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\mathrm{and}{\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right] \\ \Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}} \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ \Rightarrow \frac{1-x^2}{x^2}=\frac{4}{5} \\ \Rightarrow 5-5x^2=4x^2 \\ \Rightarrow 9x^2=5 \\ \Rightarrow x=\pm \frac{\sqrt{5}}{3} \end{gathered}$$

(ix)
$$\begin{gathered} {\tan }^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2}{\tan }^{-1}\left(x\right)=0 \\ \Rightarrow {\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}\left(x\right) \\ \Rightarrow {\tan }^{-1}1-{\tan }^{-1}x=\frac{1}{2}{\tan }^{-1}\left(x\right)\left[\because {\tan }^{-1}1-{\tan }^{-1}x={\tan }^{-1}\left(\frac{1-x}{1+x}\right)\right] \\ \Rightarrow {\tan }^{-1}1=\frac{3}{2}{\tan }^{-1}\left(x\right) \\ \Rightarrow \frac{\mathrm{\pi }}{4}=\frac{3}{2}{\tan }^{-1}\left(x\right) \\ \Rightarrow \frac{\mathrm{\pi }}{6}={\tan }^{-1}\left(x\right) \\ \Rightarrow x=\frac{1}{\sqrt{3}} \end{gathered}$$

(x)

$$\begin{gathered} \Rightarrow {\cot }^{-1}\left(x\right)-{\cot }^{-1}\left(x+2\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\tan }^{-1}\left(\frac{1}{x}\right)+{\cot }^{-1}\left(\frac{1}{x+2}\right)=\frac{\mathrm{\pi }}{12}\left[\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x}\right] \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+2}}}{1+{\displaystyle \frac{1}{x\left(x+2\right)}}}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{2}{x\left(x+2\right)}}}{{\displaystyle \frac{x^2+2x+1}{x\left(x+2\right)}}}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\tan }^{-1}\left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\tan \frac{\mathrm{\pi }}{12} \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\tan \left(\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{3}}-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{3}}\times \tan {\displaystyle \frac{\mathrm{\pi }}{4}}} \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \Rightarrow \left(\frac{2}{{\displaystyle x^2+2x+1}}\right)=\frac{2}{{\left(\sqrt{3}+1\right)}^2} \\ \Rightarrow \frac{1}{{\left({\displaystyle x+1}\right)}^2}=\frac{1}{{\left(\sqrt{3}+1\right)}^2} \\ \Rightarrow x+1=\sqrt{3}+1 \\ \Rightarrow x=\sqrt{3} \end{gathered}$$


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

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

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

$$\begin{gathered} \therefore {\tan }^{-1}\left(x+2\right)+{\tan }^{-1}\left(x-2\right)={\tan }^{-1}\frac{8}{79} \\ \Rightarrow {\tan }^{-1}\left(\frac{x+2+x-2}{1-\left(x+2\right)\times \left(x-2\right)}\right)={\tan }^{-1}\frac{8}{79} \\ \Rightarrow \frac{2x}{1-x^2+4}=\frac{8}{79} \\ \Rightarrow \frac{x}{5-x^2}=\frac{4}{79} \\ \Rightarrow 79x=20-4x^2 \\ \Rightarrow 4x^2+79x-20=0 \\ \Rightarrow 4x^2+80x-x-20=0 \\ \Rightarrow \left(4x-1\right)\left(x+20\right)=0 \\ \Rightarrow x=\frac{1}{4}\mathrm{or}-20 \\ \therefore x=\frac{1}{4}\left[\because x>0\right] \end{gathered}$$

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

$$\begin{gathered} \therefore {\tan }^{-1}\left(\frac{x}{2}\right)+{\tan }^{-1}\left(\frac{x}{3}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{x}{2}}+{\displaystyle \frac{x}{3}}}{1-{\displaystyle \frac{x}{2}}\times {\displaystyle \frac{x}{3}}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{5x}{6}}}{{\displaystyle \frac{6-x^2}{6}}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{5x}{6-x^2}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{5x}{6-x^2}=1 \\ \Rightarrow 5x=6-x^2 \\ \Rightarrow x^2+5x-6=0 \\ \Rightarrow \left(x-1\right)\left(x+6\right)=0 \\ \Rightarrow x=1\left[\because 0<x<\sqrt{6}\right] \end{gathered}$$