√1+sinA/1-sinA = sec A+ Rana prove that
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√1-sin A / 1+sin A
= √(1-sin A) / (1+sin A ) x √(1-sin A) / (1-sin A )
= √(1-sin A)2 / (1-sin2 A )
= √(1-sin A)2 / cos2 A
= (1-sin A) / cos A
= (1/cos A - sin A/cos A)
= sec A - tan A.
√1-sin A / 1+sin A = sec A - tan A.
= √(1-sin A) / (1+sin A ) x √(1-sin A) / (1-sin A )
= √(1-sin A)2 / (1-sin2 A )
= √(1-sin A)2 / cos2 A
= (1-sin A) / cos A
= (1/cos A - sin A/cos A)
= sec A - tan A.
√1-sin A / 1+sin A = sec A - tan A.
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see the picture. .................
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