Question

Prove that secA (1- SinA) (secA + tanA) =1
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Answer 1

Answer:

hence proved ,

secA ( 1-sinA) (secA+tanA) =1

Step-by-step explanation:

secA(1-sinA)(secA+tanA)=1

LHS =secA(1-sinA)(secA+tanA)

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

=(1-sinA)( (1+sinA)/(cos²A)

=(1-sin²A)/(cos²A)=cos²A/cos²A=1 = RHS

Answer 2

LHS

$secA(1-sinA)(secA+tanA) \\ \\ = \frac{1}{cosA} (1-sinA)(secA+tanA) \\ \\ = ( \frac{1}{cosA} - \frac{ sinA}{cosA} )(secA+tanA) \\ \\ = (secA-tanA)(secA+tanA) \\ \\ = {sec}^{2} A - {tan}^{2} A\\ \\ = 1$

RHS

=1

LHS=RHS