sinA + cosA =

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Solution

The correct option is D

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\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{2 {sin}^{2} A - 1}

\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n^{2} A - 1}

\frac{sin (A + B) + sin (A - B)}{cos (A + B) + cos (A - B)} = tan A

sin (A + B) + sin (A - B) = sin A

cos (A + B) + cos (A - B) = cos A

\frac{1 + cos A - sin A}{1 + cos A + sin A} = sec A - tan A = \frac{1 - sin A}{cos A} = \sqrt{\frac{1 - sin A}{1 + sin A}}

\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + sin A}{cos A}

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