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

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