Question

prove that sin A by secA + tanA -1 + cosA by cosecA+ cotA -1 = 1
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Answer 1

Step-by-step explanation:

L.H.S.

= sin A/(sec A + tan A – 1) + cos A/(cosec A + cot A – 1)

= [sin A/(1/cos A + sin A/cos A) – 1] + [cos A/(1/sin A + cos A/sin A) – 1]

= [sin A/(1 + sin A – cos A)/cos A] + [cos A/(1 + cos A – sin A)/sin A]

= sin A.cos A/1 + sin A – cos A + sin A.cos A/1 + cos A – sin A

= sin A. cos A (1 + cos A – sin A + 1 + sin A – cos A)/[1 + (sin A – cos A)][1 – (sin A – cos A)]

= 2 sin A.cos A/(1)2 – (sin A – cos A)2

= 2 sin A.cos A/(1 – (sin2 A + cos2 A – 2 sin A.cos A)

= 2 sin A.cos A/1 – 1 + 2 sin A.cos A

= 2/2 = 1 = R.H.S.

Hence proved.