prove that (1+tanA-secA) (1+tanA+secA) = 2tanA
Answers
Answered by
1
Answer
L.H.S.=[secA+(tanA−1)]
[secA−(tanA−1)]−2tanA
=sec
2
A−(tanA−1)
2
−2tanA
sec
2
A−tan
2
A−1+2tanA−2tanA
=1−1=0=R.H.S.
Answered by
1
Answer:
(1+tanA-secA)(1+tanA+secA)
Step-by-step explanation:
(1+sinA/cosA-1/cos)(1+sinA/cosA+1/cosA)
=(cos^2+cosA.sinA+cosA+sinA.cosA+sin^2+sinA-cosA-1)/cos^2
=2sinAcosA/cos^2
2tanA
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