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ID#6718 HSC Higher Math 2nd MCQ (Barisal 2025)

cotθ cot 3θ = 1 এর সাধারণ সমাধান কোনটি?
ক) (2n + 1)π/8
খ) (2n + 1)π/2
গ) (4n + 1)π/8
ঘ) (4n + 1)π/2

ব্যাখ্যা

দেওয়া আছে $\cot \theta \cot 3\theta = 1$
আমরা জানি $\cot \theta = \frac{1}{\tan \theta}$।
সুতরাং, $\frac{1}{\tan \theta} \cot 3\theta = 1$
$\cot 3\theta = \tan \theta$
আমরা জানি $\tan \theta = \cot (\frac{\pi}{2} - \theta)$।
সুতরাং, $\cot 3\theta = \cot (\frac{\pi}{2} - \theta)$
সাধারণ সমাধান হিসেবে, $3\theta = n\pi + (\frac{\pi}{2} - \theta)$, যেখানে $n \in \mathbb{Z}$।
$3\theta + \theta = n\pi + \frac{\pi}{2}$
$4\theta = n\pi + \frac{\pi}{2}$
$4\theta = \frac{2n\pi + \pi}{2}$
$\theta = \frac{(2n + 1)\pi}{8}$।
Resource Details
Exam HSC
Subject Higher Math 2nd paper
Chapter 7
Board Barisal
Year 2025

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