\( \sin(\tan^{-1} \frac{1}{2} + \cot^{-1}3) \) এর মান নিচের কোনটি?

Another Explanation (5):

সমাধান:

প্রথমে, আমরা দিতে \( \sin \left( \tan^{-1} \frac{1}{2} + \cot^{-1} 3 \right) \) এই সমীকরণটি বিবেচনা করব। ধরি: \[ A = \tan^{-1} \frac{1}{2} \] \[ B = \cot^{-1} 3 \] অর্থাৎ, \[ \tan A = \frac{1}{2} \] এবং, \[ \cot B = 3 \Rightarrow \tan B = \frac{1}{3} \] আমরা জানি যে, \[ \sin (A + B) = \sin A \cos B + \cos A \sin B \] এখন, \(\sin A, \cos A, \sin B, \cos B\) নির্ণয় করি। --- **ধাপ 1:** \(\tan A = \frac{1}{2}\) একটি ত্রিভুজে, \[ \text{Opposite} = 1, \quad \text{Adjacent} = 2 \] অতএব, \[ \text{Hypotenuse} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \] সুতরাং, \[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{5}} \] \[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2}{\sqrt{5}} \] --- **ধাপ 2:** \(\tan B = \frac{1}{3}\) একটি ত্রিভুজে, \[ \text{Opposite} = 1, \quad \text{Adjacent} = 3 \] অতএব, \[ \text{Hypotenuse} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \] সুতরাং, \[ \sin B = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1}{\sqrt{10}} \] \[ \cos B = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{\sqrt{10}} \] --- **ধাপ 3:** \(\sin (A + B)\) হিসাব করি: \[ \sin (A + B) = \sin A \cos B + \cos A \sin B \] প্রতিস্থাপন করি: \[ = \left( \frac{1}{\sqrt{5}} \times \frac{3}{\sqrt{10}} \right) + \left( \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{10}} \right) \] সাধারিত করি: \[ = \frac{3}{\sqrt{5} \times \sqrt{10}} + \frac{2}{\sqrt{5} \times \sqrt{10}} \] দুটি ভগ্নাংশের সাধারণ মূল: \[ \sqrt{5} \times \sqrt{10} = \sqrt{5 \times 10} = \sqrt{50} = 5 \sqrt{2} \] অতএব, \[ \sin (A + B) = \frac{3}{5 \sqrt{2}} + \frac{2}{5 \sqrt{2}} = \frac{3 + 2}{5 \sqrt{2}} = \frac{5}{5 \sqrt{2}} = \frac{1}{\sqrt{2}} \] --- **অতএব,** \[ \boxed{\sin \left( \tan^{-1} \frac{1}{2} + \cot^{-1} 3 \right) = \frac{1}{\sqrt{2}}} \] **উত্তর: \(\frac{1}{\sqrt{2}}\)**