Question

(sec A + tan A) (1 − sin A) =

A. sec A 

B. sin A 

C. cosec A 

D. cos A

Answer 1

To find: (sec A + tan A) (1 − sin A)

Consider (sec A + tan A) (1 − sin A)

We know that sec A = \(\frac{1}{cos A}\) and tan A = \(\frac{sinA}{cosA}\) 

⇒ (sec A + tan A) (1 − sin A) = \(\Big(\frac{1}{cosA}+\frac{sinA}{cosA}\Big)\)(1-sinA)

= \(\Big(\frac{1+sinA}{cosA}\Big)\) (1-sinA)

∵ (a + b) (a – b) = a² – b²

∴ (sec A + tan A) (1 − sin A) = \(\Big(\frac{1+sinA}{cosA}\Big)\) (1-sinA) = \(\frac{1-sin^2A}{cosA}\) 

Also, sin²A + cos²A = 1 

⇒ 1 – sin²A = cos²A

⇒ (sec A + tan A) (1 − sin A) = \(\frac{1-sin^2A}{cosA}\) = \(\frac{cos^2A}{cosA}\) = cosA