prove that
(1+tanA-secA)×(1+tanA+secA)
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tanA−secA+1
tanA+secA−1
=
cosA
1+sinA
Taking L.H.S.-
tanA−secA+1
tanA+secA−1
=
tanA−secA+1
(tanA+secA)−(sec
2
A−tan
2
A)
[∵1+tan
2
A=sec
2
A]
=
tanA−secA+1
(tanA+secA)−(secA+tanA)(secA−tanA)
=
tanA−secA+1
(tanA+secA)(1−(secA−tanA))
=
tanA−secA+1
(tanA+secA)(1−secA+tanA)
=tanA+secA
=
cosA
sinA
+
cosA
1
=
cosA
1+sinA
= R.H.S.
Hence proved.
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