Question

Prove that 1-sinA/1+SinA =(secA-tanA)^2​

Answer 1

Answer:

Proved below.

Step-by-step explanation:

(1 - sinA)/(1 + sinA)

Rationalize (1 + sinA) by multiplying the numerator and denominator with (1 - sinA)

⇒(1 - sinA)/(1 + sinA)

  = (1 - sinA)*(1 - sinA)/(1 + sinA)*(1 - sinA)

  = (1 - sinA)²/(1 - sin²A)     [∵sin²A + cos²A = 1 ⇒ cos²A = 1 - sin²A]

  = (1 - sinA)²/ cos²A

  = (1 - sinA/cosA)²

  = (1/cosA - sinA/cosA)²       [∵1/cosA = secA and sinA/cosA = tanA]

  = (secA - tanA)²

∴(1 - sinA)/(1 + sinA) = (secA + tanA)²

Hence proved