Question

secA (1-sina) (secA + tanA)=1​

Answer 1

Step-by-step explanation:

please please please pleae mark it as brainliest

Answer 2

PLEASE MARK AS BRAINLIEST

100% Correct answer.

secA (1-sinA) (secA + tan A) 

secA (1-sinA) (secA + tan A) = (secA-sinA×secA)(secA+tanA)

secA (1-sinA) (secA + tan A) = (secA-sinA×secA)(secA+tanA)=(secA-tanA) (secA+tanA)  as secA=1/cosA and sinA/cosA=tanA

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

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.

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.

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.Hope this helps you .