Question

(sec A + tan A) (1 – sin A) = ?​

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.

Answer 2

Answer:

CosA

Step-by-step explanation:

secA = 1/cosA

tanA = sinA/cosA

secA + tanA = (1+sinA)/cosA

(secA + tanA)*(1 - sinA) = ((1 + sinA) / cosA ) * (1 - sinA)

= (1 - sin^2A)/cosA

= cos^2A/cosA

= cosA