secA(1-sinA)(secA+tana)=1 prove that
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Answer:
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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Answered by
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secA ( 1 - sinA ) (secA + tan A)
⇒ (secA - sinA × secA)( secA + tanA )
⇒ (secA - sinA/cosA)( secA + tanA )
⇒ ( secA - tanA) ( secA + tanA)
⇒ sec²A - tan²A
⇒ 1
Proved.
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