যদি, 2tan^-1(1/5)-tan^-1(1/3)=tan^-1x হয় তবে x এর মান কত?

প্রথমে, আমাদের দেওয়া সমীকরণ হলো:

\(2 \tan^{-1} \left(\frac{1}{5}\right) - \tan^{-1} \left(\frac{1}{3}\right) = \tan^{-1} x\)

এখন, \(2 \tan^{-1} \left(\frac{1}{5}\right)\) এর জন্য, আমরা ব্যবহার করব ত্রিগনোমেট্রিক সূত্র:

\(2 \tan^{-1} a = \tan^{-1} \left( \frac{2a}{1 - a^2} \right)\)

অর্থাৎ,

\(2 \tan^{-1} \left(\frac{1}{5}\right) = \tan^{-1} \left( \frac{2 \times \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2} \right) = \tan^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right)\)

গণনা করি:

\(= \tan^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan^{-1} \left( \frac{2}{5} \times \frac{25}{24} \right) = \tan^{-1} \left( \frac{50}{120} \right) = \tan^{-1} \left( \frac{5}{12} \right)\)

এখন, সমীকরণটি হবে:

\(\tan^{-1} \left( \frac{5}{12} \right) - \tan^{-1} \left( \frac{1}{3} \right) = \tan^{-1} x\)

আমরা জানি,

\(\tan^{-1} A - \tan^{-1} B = \tan^{-1} \left( \frac{A - B}{1 + AB} \right)\)

অতএব,

\(x = \frac{\frac{5}{12} - \frac{1}{3}}{1 + \frac{5}{12} \times \frac{1}{3}}\)

\(\frac{5}{12} - \frac{1}{3} = \frac{5}{12} - \frac{4}{12} = \frac{1}{12}\)

\(1 + \frac{5}{12} \times \frac{1}{3} = 1 + \frac{5}{36} = \frac{36}{36} + \frac{5}{36} = \frac{41}{36}\)

\(x = \frac{\frac{1}{12}}{\frac{41}{36}} = \frac{1}{12} \times \frac{36}{41} = \frac{36}{12 \times 41} = \frac{3}{41}\)

অতএব,

উত্তর: \(x = \frac{3}{41}\)