Question

Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Answer 1

1−sinA1+sinA=(1−sinA)(1−sinA)(1+sinA)(1−sinA)

=(1−sinA)21−sin2A

=(1−2sinA+sin2A)cos2A

=(1cos2A−2sinAcos2A+sin2Acos2A)

=(sec2A−2.1cosA.sinAcosA+tan2A)

=(sec2A−2secA.tanA+tan2A)

=(secA−tanA)2

Hence proved.

Note: The question should be (1−sinA1+sinA)=(secA−tanA)2

Answer 2

Answer:

Step-by-step explanation:

⇒sec A (1 – sin A)(sec A + tan A) = 1.

⇒sec A (1 – sin A)($\frac{1}{cosA} +\frac{sinA}{cosA}$)

⇒sec A (1 – sin A)($\frac{1+sinA}{cosA}$)

⇒sec A (1 – sin A)   sec A (1 + sin A)

using (a+b)(a-b)=(a²-b²)

⇒sec A (1-sin²A)

as we know 1-sin²A = cosA

⇒sec A (cos A)

⇒1