Question

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

Answer 1

secA(1-sinA)(secA+tanA)
= (1/cosA)(1-sinA)(1/cosA+sinA/cosA)
= (1/cos²A)(1-sinA)(1+sinA)
= (1/cos²A)(1-sin²A)
= (1/cos²A)(cos²A)
= 1
= RHS

Answer 2

secA (1-sinA) (secA + tan A) 
= (secA-sinA×secA)(secA+tanA)
=(secA-tanA) (secA+tanA)  as secA=1/cosA and sinA/cosA=tanA
= (sec²A-tan²A) as (a+b)(a-b)=a²-b²
=sec²A-tan²A=1 from identity.
Hence proved.

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