cot^(-1)3+cosec^-1√5=? 

Explanation:

Another Explanation (5): bài toán: \(cot^{-1}3+cosec^{-1}\sqrt{5}=?\) giải: মনে করি, \(cot^{-1}3 = A\) এবং \(cosec^{-1}\sqrt{5} = B\) তাহলে, \(cotA = 3\) এবং \(cosec B = \sqrt{5}\) এখন, \(tanA = \frac{1}{cotA} = \frac{1}{3}\) এবং, \(sinB = \frac{1}{cosec B} = \frac{1}{\sqrt{5}}\) আমরা জানি, \(sin^2B + cos^2B = 1\) সুতরাং, \(cos^2B = 1 - sin^2B = 1 - (\frac{1}{\sqrt{5}})^2 = 1 - \frac{1}{5} = \frac{4}{5}\) তাহলে, \(cosB = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\) এখন, \(tanB = \frac{sinB}{cosB} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2}\) এখন, আমরা \(tan(A+B)\) এর মান বের করি। \(tan(A+B) = \frac{tanA + tanB}{1 - tanA \cdot tanB} = \frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} \cdot \frac{1}{2}} = \frac{\frac{2+3}{6}}{1 - \frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1\) যেহেতু, \(tan(A+B) = 1\) সুতরাং, \(A+B = tan^{-1}(1) = \frac{\pi}{4}\) 🥳 অতএব, \(cot^{-1}3+cosec^{-1}\sqrt{5} = \frac{\pi}{4}\)