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