Another Explanation (5): প্রশ্ন: \( 2 \tan^{-1} \left( \frac{1}{3} \right) \) এর সমতুল্য মান কোনটি?
নিচের অপশনগুলো হলো:
i) \( \sin^{-1} \left( \frac{3}{5} \right) \)
ii) \( \cos^{-1} \left( \frac{4}{5} \right) \)
iii) \( \tan^{-1} \left( \frac{2}{3} \right) \)
উত্তর: "i ও ii"
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প্রথমে, ধরা যাক \( \theta = \tan^{-1} \left( \frac{1}{3} \right) \)। তাহলে,
\[
\tan \theta = \frac{1}{3}
\]
এখন, \( 2 \theta = 2 \tan^{-1} \left( \frac{1}{3} \right) \), অর্থাৎ,
\[
\text{আমরা জানি} \quad \tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}
\]
অতএব,
\[
\tan 2 \theta = \frac{2 \times \frac{1}{3}}{1 - \left( \frac{1}{3} \right)^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{18}{24} = \frac{3}{4}
\]
অর্থাৎ,
\[
\tan 2 \theta = \frac{3}{4}
\]
এখন, \( 2 \theta \) এর মান \( \alpha \) হিসেবে ধরে নিই, যেখানে
\[
\tan \alpha = \frac{3}{4}
\]
আমরা জানি,
\[
\text{সুতরাং,} \quad \alpha = \tan^{-1} \left( \frac{3}{4} \right)
\]
এখন, একটি রিলেশন আছে:
\[
\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}}
\]
তাই, প্রথমে হাইপোটেনুসের মান নির্ণয় করি:
\[
\text{Hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
অতএব,
\[
\sin \alpha = \frac{3}{5}
\]
\[
\cos \alpha = \frac{4}{5}
\]
এখন,
\[
\alpha = \tan^{-1} \left( \frac{3}{4} \right)
\]
তাই,
\[
2 \theta = \alpha \Rightarrow \theta = \frac{\alpha}{2}
\]
এবং,
\[
\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}
\]
\[
\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}
\]
তাই,
\[
\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}
\]
এবং,
\[
\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{4}{5}}{2}} = \sqrt{\frac{\frac{9}{5}}{2}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}
\]
তাহলে,
\[
\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = 2 \times \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = 2 \times \frac{3}{10} = \frac{6}{10} = \frac{3}{5}
\]
আর,
\[
\cos \alpha = \cos^2 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2} = \left(\frac{3}{\sqrt{10}}\right)^2 - \left(\frac{1}{\sqrt{10}}\right)^2 = \frac{9}{10} - \frac{1}{10} = \frac{8}{10} = \frac{4}{5}
\]
অর্থাৎ,
\[
\sin \alpha = \frac{3}{5}, \quad \cos \alpha = \frac{4}{5}
\]
এবং, \( \alpha = \sin^{-1} \left( \frac{3}{5} \right) \) অথবা \( \cos^{-1} \left( \frac{4}{5} \right) \)
কিন্তু, কারণ \( 2 \theta = \alpha \), তাই:
\[
\theta = \frac{\alpha}{2}
\]
অর্থাৎ,
\[
2 \tan^{-1} \left( \frac{1}{3} \right) = \alpha \Rightarrow \tan \alpha = \frac{3}{4}
\]
তাহলে, \( \alpha = \tan^{-1} \left( \frac{3}{4} \right) \).
তাই,
\[
2 \tan^{-1} \left( \frac{1}{3} \right) = \sin^{-1} \left( \frac{3}{5} \right) \quad \text{ও} \quad \cos^{-1} \left( \frac{4}{5} \right)
\]
এবং, অপশনে (i) ও (ii) উভয়ই সঠিক।
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সুতরাং, সঠিক উত্তর হলো: **"i ও ii"**
