Correct option is AGiven:1+sin A1−sin A\sqrt\frac{1+\text{sin\space\text{A}}}{1-\text{sin}\space\text{A}}1−sin A1+sin AFormula Used:sin2A + cos2A = 1secA=1cosA\sec A=\frac 1{\cos A}secA=cosA1 , tan A =sinAcosA\frac{\sin A}{\cos A}cosAsinASolution:1+sin A1−sin A\sqrt\frac{1+\text{sin\space\text{A}}}{1-\text{sin}\space\text{A}}1−sin A1+sin A=1+sin A1−sin A×1+sin A1+sin A\sqrt{\frac{1+\text{sin\space\text{A}}}{1-\text{sin}\space\text{A}} \times\frac{1+\text{sin\space\text{A}}}{1+\text{sin}\space\text{A}} }1−sin A1+sin A×1+sin A1+sin A= (1+sin A)21−sin2A\sqrt\frac{(1+\text{sin A})^2}{1-\sin^2 A}1−sin2A(1+sin A)2= 1+sin AcosA\frac{1+\text{sin\space\text{A}}}{\cos A}cosA1+sin A= 1cosA+sinAcosA\frac 1{\cos A}+\frac{\sin A}{\cos A}cosA1+cosAsinA= sec A + tan A