Question

If tan⁻¹ (cot θ) = 2 θ, then θ =

(a) $\pm \frac{\mathrm{\pi }}{3}$

(b) $\pm \frac{\mathrm{\pi }}{4}$

(c) $\pm \frac{\mathrm{\pi }}{6}$

(d) none of these

Answer 1

(c) $\pm \frac{\mathrm{\pi }}{6}$


$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ {\tan }^{-1}\left(cot\theta \right)=2\theta \\ \Rightarrow \tan 2\theta =cot\theta \\ \Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{1}{\tan \theta } \\ \Rightarrow 2{\tan }^2\theta =1-{\tan }^2\theta \\ \Rightarrow 3{\tan }^2\theta =1 \\ \Rightarrow {\tan }^2\theta =\frac{1}{3} \\ \Rightarrow \tan \theta =\pm \frac{1}{\sqrt{3}} \\ \therefore \theta =\pm \frac{\mathrm{\pi }}{6} \end{gathered}$$