$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Solve} \mathrm{for} \mathrm{x},2{\tan }^{-1}\mathrm{x} (\cos \mathrm{x}) = {\tan }^{-1} \left(2\mathrm{cosecx}\right). \\ \left(\mathrm{b}\right) \mathrm{Solve} \mathrm{for} \mathrm{x},{\tan }^{-1}(\mathrm{x} – 1) + {\tan }^{-1} \mathrm{x} + {\tan }^{-1}(\mathrm{x} + 1) = {\tan }^{-1} 3\mathrm{x} \\ \mathrm{OR} \end{gathered}$$

(a) Prove that 

${\tan }^{-1}\left(\frac{6\mathrm{x}-8{\mathrm{x}}^3}{1-12{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{4\mathrm{x}}{1-4{\mathrm{x}}^2}\right)={\tan }^{-1}2\mathrm{x}; \left|2\mathrm{x}\right|<\frac{1}{\sqrt{3}}$

$\left(\mathrm{b}\right) \mathrm{Prove} \mathrm{that}3 {\sin }^{-1}\mathrm{x} = {\sin }^{-1}(3\mathrm{x} – 4{\mathrm{x}}^3), \mathrm{x}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right].$

(a)
Given equation is 

$2 {\tan }^{-1} (\cos \mathrm{x}) = {\tan }^{-1} (2 \mathrm{cosec} \mathrm{x})$

$$\begin{gathered} \Rightarrow {\tan }^{-1}\left(\frac{2 \cos \mathrm{x}}{1-{\cos }^2 \mathrm{x}}\right)={\tan }^{-1}\left(\frac{2}{\sin \mathrm{x}}\right) \\ \left[\because 2 {\tan }^{-1} \mathrm{x}={\tan }^{-1} \left(\frac{2\mathrm{x}}{1-{\mathrm{x}}^2}\right); -1<\mathrm{x}<1\right] \\ \Rightarrow \frac{2 \cos \mathrm{x}}{{\sin }^2 \mathrm{x}} =\frac{2}{\sin \mathrm{x}} [\because 1-{\cos }^2 \mathrm{x}={\sin }^2 \mathrm{x}] \dots \left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \sin \mathrm{x} \cos \mathrm{x} – {\sin }^2 \mathrm{x} = 0 \\ \Rightarrow \sin \mathrm{x} (\cos \mathrm{x} – \sin \mathrm{x}) = 0 \\ \Rightarrow \sin \mathrm{x} = 0 \mathrm{or} \cos \mathrm{x} = \sin \mathrm{x} \\ \Rightarrow \sin \mathrm{x} = \sin 0 \\ \mathrm{or} \cot \mathrm{x} = 1 = \cot \left(\frac{\mathrm{\pi }}{4}\right) \\ \therefore \mathrm{x} = 0 \mathrm{or} \left(\frac{\mathrm{\pi }}{4}\right) \\ \mathrm{But} \mathrm{here} \mathrm{at} \mathrm{x} = 0, \mathrm{the} \mathrm{given} \mathrm{equation} \mathrm{does} \mathrm{not} \mathrm{exist}. \\ \mathrm{Hence}, \mathrm{x}=\frac{\mathrm{\pi }}{4} \mathrm{is} \mathrm{the} \mathrm{only} \mathrm{solution}. \end{gathered}$$

(b)
Given, 

$$\begin{gathered} {\tan }^{-1} (\mathrm{x} – 1) + {\tan }^{-1} \mathrm{x} + {\tan }^{-1} (\mathrm{x} + 1) = {\tan }^{-1} 3\mathrm{x} \\ \Rightarrow {\tan }^{-1} (\mathrm{x} – 1) + {\tan }^{-1} (\mathrm{x} + 1) ={\tan }^{-1} 3\mathrm{x}-{\tan }^{-1} \mathrm{x} \end{gathered}$$

$$\begin{gathered} \Rightarrow {\tan }^{-1}\left(\frac{\mathrm{x}-1+\mathrm{x}+1}{1-(\mathrm{x}-1) (\mathrm{x}+1)}\right)={\tan }^{-1} \left(\frac{3\mathrm{x}-\mathrm{x}}{1+3\mathrm{x}\left(\mathrm{x}\right)}\right) \\ \left[\begin{array}{c}\because {\tan }^{-1} \mathrm{x}+{\tan }^{-1} \mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right), \mathrm{xy}<1\\ \mathrm{and} {\tan }^{-1}\mathrm{x}-{\tan }^{-1} \mathrm{y}={\tan }^{-1} \left(\frac{\mathrm{x}-\mathrm{y}}{1+\mathrm{xy}}\right), \mathrm{xy}>-1\end{array}\right] \\ \Rightarrow {\tan }^{-1}\left(\frac{2\mathrm{x}}{1-({\mathrm{x}}^2-1)}\right)={\tan }^{-1}\left(\frac{2\mathrm{x}}{1+3{\mathrm{x}}^2}\right) \\ \Rightarrow \frac{2\mathrm{x}}{2-{\mathrm{x}}^2}=\frac{2\mathrm{x}}{1+3{\mathrm{x}}^2} \end{gathered}$$

$$\begin{gathered} \Rightarrow 2\mathrm{x}(1 + 3{\mathrm{x}}^2) = 2\mathrm{x}(2 – {\mathrm{x}}^2) \\ \Rightarrow 2\mathrm{x}[1 + 3{\mathrm{x}}^2 – (2 – {\mathrm{x}}^2\left)\right] = 0 \\ \Rightarrow \mathrm{x}(4{\mathrm{x}}^2 – 1) = 0 \Rightarrow \mathrm{x} = 0 \mathrm{or} 4{\mathrm{x}}^2 – 1 = 0 \\ \therefore \mathrm{x} = 0 \mathrm{or} \mathrm{x} = \pm \frac{1}{2} \end{gathered}$$

                               OR

(a)

$$\begin{gathered} \mathrm{To} \mathrm{prove}, {\tan }^{-1}\left(\frac{6\mathrm{x}-8{\mathrm{x}}^2}{1-12{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{4\mathrm{x}}{1-4{\mathrm{x}}^2}\right) \\ ={\tan }^{-1} 2\mathrm{x}; \left|2\mathrm{x}\right|<\frac{1}{\sqrt{3}} \\ \mathrm{We} \mathrm{consider}, \\ \mathrm{LHS}={\tan }^{-1}\left(\frac{6\mathrm{x}-8{\mathrm{x}}^3}{1-12{\mathrm{x}}^2}\right)-{\tan }^{-1}\left(\frac{4\mathrm{x}}{1-4{\mathrm{x}}^2}\right) \\ ={\tan }^{-1}\left[\frac{\left({\displaystyle \frac{6\mathrm{x}-8{\mathrm{x}}^2}{1-12{\mathrm{x}}^2}}\right)-\left({\displaystyle \frac{4\mathrm{x}}{1-4{\mathrm{x}}^2}}\right)}{1+\left({\displaystyle \frac{6\mathrm{x}-8{\mathrm{x}}^3}{1-12{\mathrm{x}}^2}}\right) \left({\displaystyle \frac{4\mathrm{x}}{1-4{\mathrm{x}}^2}}\right)}\right] \\ \left[\because {\tan }^{-1} \mathrm{x}-{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{1+\mathrm{xy}}\right); \mathrm{xy}>-1\right] \\ ={\tan }^{-1}\left[\frac{{\displaystyle \frac{(6\mathrm{x}-8{\mathrm{x}}^3) (1-4{\mathrm{x}}^2)-4\mathrm{x}(1-12{\mathrm{x}}^2)}{(1-12{\mathrm{x}}^2) (1-4{\mathrm{x}}^2)}}}{{\displaystyle \frac{(1-12{\mathrm{x}}^2) (1-4{\mathrm{x}}^2)+(6\mathrm{x}-8{\mathrm{x}}^3) \left(4\mathrm{x}\right)}{(1-12{\mathrm{x}}^2) (1-4{\mathrm{x}}^2)}}}\right] \\ ={\tan }^{-1}\left(\frac{6\mathrm{x}-24{\mathrm{x}}^3-8{\mathrm{x}}^3+32{\mathrm{x}}^5-4\mathrm{x}+48{\mathrm{x}}^3}{1-4{\mathrm{x}}^2-12{\mathrm{x}}^2+48{\mathrm{x}}^4+24{\mathrm{x}}^2-32{\mathrm{x}}^4}\right) \\ ={\tan }^{-1}\left(\frac{2\mathrm{x}+16{\mathrm{x}}^3+32{\mathrm{x}}^5}{16{\mathrm{x}}^4+8{\mathrm{x}}^2+1}\right) \\ ={\tan }^{-1}\left[\frac{2\mathrm{x}(16{\mathrm{x}}^4+8{\mathrm{x}}^2+1)}{(16{\mathrm{x}}^4+8{\mathrm{x}}^2+1)}\right]={\tan }^{-1} 2\mathrm{x}=\mathrm{RHS} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{b}\right)\mathrm{Consider}, \mathrm{RHS} = {\sin }^{-1} (3\mathrm{x} – 4{\mathrm{x}}^3) \dots \dots \left(\mathrm{i}\right) \\ \mathrm{Let} \mathrm{x} = \sin \mathrm{\theta }, \\ \mathrm{then} \mathrm{\theta } = {\sin }^{-1} \mathrm{x} \\ \mathrm{Now}, \mathrm{from} \mathrm{Eq}. \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \mathrm{RHS} = {\sin }^{-1} (3 \sin \mathrm{\theta } – 4 {\sin }^3 \mathrm{\theta }) \\ = {\sin }^{-1} (\sin 3 \mathrm{\theta }) [\because \sin 3\mathrm{A} = 3\sin \mathrm{A} – 4{\sin }^3 \mathrm{A}] \\ = 3\mathrm{\theta } \\ = 3 {\sin }^{-1} \mathrm{x} [\because \mathrm{\theta } = {\sin }^{-1} \mathrm{x}] \\ = \mathrm{LHS} \\ \mathrm{Hence} \mathrm{Proved}. \end{gathered}$$

$\left[\begin{array}{c}\because {\sin }^{-1} (\mathrm{sin\theta }=\mathrm{\theta }, \forall \in \left[\frac{-\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2}\right]\\ \mathrm{and} \mathrm{here} -\frac{1}{2}\le \mathrm{x}\le \frac{1}{2}\Rightarrow \frac{\mathrm{\pi }}{6}\le {\sin }^{-1} \mathrm{x}\le \frac{\mathrm{\pi }}{6}\end{array}\right]$