tan2θtanθ=1 এর সমাধান কোনটি?

Explanation:

Another Explanation (5): bài
দেওয়া আছে, \( \tan 2\theta \tan \theta = 1 \)
\( \Rightarrow \tan 2\theta = \frac{1}{\tan \theta} \)
\( \Rightarrow \tan 2\theta = \cot \theta \)
\( \Rightarrow \tan 2\theta = \tan \left( \frac{\pi}{2} - \theta \right) \)
আমরা জানি, \( \tan x = \tan y \) হলে, \( x = n\pi + y \) হয়। এখানে, \( n \in \mathbb{Z} \)
অতএব, \( 2\theta = n\pi + \frac{\pi}{2} - \theta \)
\( \Rightarrow 3\theta = n\pi + \frac{\pi}{2} \)
\( \Rightarrow \theta = \frac{n\pi}{3} + \frac{\pi}{6} \) ...(1)
আবার, \( \tan 2\theta \tan \theta = 1 \)
\( \Rightarrow \frac{\sin 2\theta}{\cos 2\theta} \cdot \frac{\sin \theta}{\cos \theta} = 1 \)
\( \Rightarrow \sin 2\theta \sin \theta = \cos 2\theta \cos \theta \)
\( \Rightarrow \cos 2\theta \cos \theta - \sin 2\theta \sin \theta = 0 \)
\( \Rightarrow \cos (2\theta + \theta) = 0 \)
\( \Rightarrow \cos 3\theta = 0 \)
আমরা জানি, \( \cos x = 0 \) হলে, \( x = (2n+1)\frac{\pi}{2} \) হয়। এখানে, \( n \in \mathbb{Z} \)
অতএব, \( 3\theta = (2n+1)\frac{\pi}{2} \)
\( \Rightarrow \theta = (2n+1)\frac{\pi}{6} \)
\( \Rightarrow \theta = \frac{2n\pi}{6} + \frac{\pi}{6} \)
\( \Rightarrow \theta = \frac{n\pi}{3} + \frac{\pi}{6} \) ...(2)
এখন, প্রদত্ত অপশন \( \theta = n\pi \pm \frac{\pi}{6} \)
যদি \( \theta = n\pi + \frac{\pi}{6} \) হয়, তবে \(\tan \theta \) এবং \(\tan 2\theta \) সংজ্ঞায়িত হতে হবে।
যদি \( \theta = n\pi - \frac{\pi}{6} \) হয়, তবে \(\tan \theta \) এবং \(\tan 2\theta \) সংজ্ঞায়িত হতে হবে।
😈😈😈 এখন, \(\tan 2\theta \tan \theta = 1 \)
\(\Rightarrow \frac{2\tan^2 \theta}{1-\tan^2 \theta}=1 \) \(\Rightarrow {2\tan^2 \theta}=1-\tan^2 \theta \) \(\Rightarrow {3\tan^2 \theta}=1 \) \(\Rightarrow {\tan^2 \theta}=\frac{1}{3} \) \(\Rightarrow {\tan \theta}=\pm \frac{1}{\sqrt3} \) \(\Rightarrow {\theta}=n\pi \pm \frac{\pi}{6} \) যেখানে n যে কোনো পূর্ণসংখ্যা।🥳🥳🥳