Question

If 3 tan^-1x + cot^-1x = π, then x equals
​(a) 0 (b) 1 (c) -1 (d) $\frac{1}{2}$

Answer 1

$$\begin{gathered} 3{\tan }^{-1}x+{\cot }^{-1}x=\mathrm{\pi } \\ \Rightarrow 2{\tan }^{-1}x+{\tan }^{-1}x+{\cot }^{-1}x=\mathrm{\pi } \\ \Rightarrow 2{\tan }^{-1}x+\frac{\mathrm{\pi }}{2}=\mathrm{\pi }\left[{\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow 2{\tan }^{-1}x=\mathrm{\pi }-\frac{\mathrm{\pi }}{2}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{4} \\ \Rightarrow x=\tan \frac{\mathrm{\pi }}{4}=1 \end{gathered}$$

Thus, the value of x is 1.

Hence, the correct answer is option (b).