sin^-1 (4/5)+cos^-1 (2/√5) = ?

Another Explanation (5): প্রশ্ন: \(\sin^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) = ?\) সমাধান: প্রথমে, ধরা যাক: \[ A = \sin^{-1}\left(\frac{4}{5}\right) \] অর্থাৎ, \[ \sin A = \frac{4}{5} \] এখন, \(\cos A\) নির্ণয় করি: \[ \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \] ধরা যাক, \[ B = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \] অর্থাৎ, \[ \cos B = \frac{2}{\sqrt{5}} \] এখন, \(\sin B\) নির্ণয় করি: \[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{2}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \] এখন, লক্ষ্য করুন: \[ A + B = \theta \] অর্থাৎ, আমরা চাই: \[ \sin A + \cos B \Rightarrow \text{তবে, } \sin \theta = \sin (A + B) \] \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \] এখন, উপাদানগুলো স্থানান্তর করি: \[ = \left(\frac{4}{5}\right) \left(\frac{2}{\sqrt{5}}\right) + \left(\frac{3}{5}\right) \left(\frac{1}{\sqrt{5}}\right) \] অঙ্কন করি: \[ = \frac{4 \times 2}{5 \times \sqrt{5}} + \frac{3 \times 1}{5 \times \sqrt{5}} = \frac{8}{5 \sqrt{5}} + \frac{3}{5 \sqrt{5}} = \frac{8 + 3}{5 \sqrt{5}} = \frac{11}{5 \sqrt{5}} \] এখন, এর মানে: \[ \sin \theta = \frac{11}{5 \sqrt{5}} \] তাই, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] \[ \cos \theta = \cos (A + B) \] \[ = \cos A \cos B - \sin A \sin B \] প্রতিটি উপাদান ব্যবহার করি: \[ = \left(\frac{3}{5}\right) \left(\frac{2}{\sqrt{5}}\right) - \left(\frac{4}{5}\right) \left(\frac{1}{\sqrt{5}}\right) = \frac{6}{5 \sqrt{5}} - \frac{4}{5 \sqrt{5}} = \frac{6 - 4}{5 \sqrt{5}} = \frac{2}{5 \sqrt{5}} \] সুতরাং, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{11}{5 \sqrt{5}}}{\frac{2}{5 \sqrt{5}}} = \frac{11}{2} \] অর্থাৎ, \[ \theta = \tan^{-1}\left(\frac{11}{2}\right) \] অতএব, \[ \boxed{\sin^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) = \tan^{-1}\left(\frac{11}{2}\right)} \]