C-09-Q097mediumsingle_mcqtanA⋅1−sin2A=?\tan A \cdot \sqrt{1 - \sin^2 A} = ?tanA⋅1−sin2A=?asinA\sin AsinAbcosA\cos AcosActanA\tan AtanAdcotA\cot AcotAব্যাখ্যাSince 1−sin2A=cosA\sqrt{1 - \sin^2 A} = \cos A1−sin2A=cosA, the product is tanA⋅cosA=sinAcosA⋅cosA=sinA\tan A\cdot\cos A = \dfrac{\sin A}{\cos A}\cdot\cos A = \sin AtanA⋅cosA=cosAsinA⋅cosA=sinA.