C-09-Q097mediumsingle_mcqtanA⋅1−sin2A=?\tan A \cdot \sqrt{1 - \sin^2 A} = ?tanA⋅1−sin2A=?asinA\sin AsinAbcosA\cos AcosActanA\tan AtanAdব্যাখ্যাSince 1−sin2A=cosA\sqrt{1 - \sin^2 A} = \cos A1−sin2A=, the product is .