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(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹$\left(\frac{1-x}{1+x}\right)-\frac{1}{2}$tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) = $\frac{\mathrm{\pi }}{12}$, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹$\left(\frac{8}{79}\right)$, x > 0
(vii) ${\tan }^{-1}\frac{x}{2}+{\tan }^{-1}\frac{x}{3}=\frac{\mathrm{\pi }}{4},0<x<\sqrt{6}$
(viii) ${\tan }^{-1}\left(\frac{x-2}{x-4}\right)+{\tan }^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\mathrm{\pi }}{4}$
(ix) ${\tan }^{-1}\left(2+x\right)+{\tan }^{-1}\left(2-x\right)={\tan }^{-1}\frac{2}{3},\mathrm{where}x<-\sqrt{3}\mathrm{or},x>\sqrt{3}$
(x) ${\tan }^{-1}\frac{x-2}{x-1}+{\tan }^{-1}\frac{x+2}{x+1}=\frac{\mathrm{\pi }}{4}$

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

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

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

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

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

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

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

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

(viii)
We know


$$\begin{gathered} \therefore {\tan }^{-1}\left(\frac{x-2}{x-4}\right)+{\tan }^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{x-2}{x-4}}+{\displaystyle \frac{x+2}{x+4}}}{1-{\displaystyle \frac{x-2}{x-4}}\times {\displaystyle \frac{x+2}{x+4}}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{{\displaystyle \frac{x^2+2x-8+x^2-2x-8}{\left(x-4\right)\left(x+4\right)}}}{{\displaystyle \frac{{\displaystyle x^2-16-x^2+4}}{{\displaystyle \left(x-4\right)\left(x+4\right)}}}}\right)=\frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{2x^2-16}{-12}=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow \frac{2x^2-16}{-12}=1 \\ \Rightarrow 2x^2-16=-12 \\ \Rightarrow 2x^2=4 \\ \Rightarrow x^2=2 \\ \Rightarrow x=\pm \sqrt{2} \end{gathered}$$${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

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

$$\begin{gathered} \therefore {\tan }^{-1}\left(x+2\right)+{\tan }^{-1}\left(x-2\right)={\tan }^{-1}\frac{2}{3} \\ \Rightarrow {\tan }^{-1}\left(\frac{2+x+2-x}{1-\left(2+x\right)\times \left(2-x\right)}\right)={\tan }^{-1}\frac{2}{3} \\ \Rightarrow \frac{4}{1-4+x^2}=\frac{2}{3} \\ \Rightarrow -6+2x^2=12 \\ \Rightarrow 2x^2=18 \\ \Rightarrow x^2=9 \\ \Rightarrow x=\pm 3 \end{gathered}$$

(x)
$$\begin{gathered} {\tan }^{-1}\frac{x-2}{x-1}+{\tan }^{-1}\frac{x+2}{x+1}=\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\tan }^{-1}\left(\frac{\left({\displaystyle \frac{x-2}{x-1}}\right)+\left({\displaystyle \frac{x+2}{x+1}}\right)}{1-\left({\displaystyle \frac{x-2}{x-1}}\right)\left({\displaystyle \frac{x+2}{x+1}}\right)}\right)=\frac{\mathrm{\pi }}{4}\left[{\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right] \\ \Rightarrow \frac{\left({\displaystyle \frac{\left(x-2\right)\left(x+1\right)+\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}}\right)}{\left({\displaystyle \frac{\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}}\right)}=\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \frac{\left(x-2\right)\left(x+1\right)+\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)-\left(x-2\right)\left(x+2\right)}=1 \\ \Rightarrow \frac{x^2-x-2+x^2+x-2}{\left(x^2-1\right)-\left(x^2-4\right)}=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{2x^2-4}{3}=1 \\ \Rightarrow 2x^2-4=3 \\ \Rightarrow 2x^2=7 \\ \Rightarrow x^2=\frac{7}{2} \\ \therefore x=\pm \sqrt{\frac{7}{2}} \end{gathered}$$