prove that tanA+ sec -1/sinA cosA-1×cosA=1+sinA/cosA
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ANSWER
(secA+tanA)
1
−
cosA
1
=
cosA
1
−
(secA−tanA)
1
∴
(secA+tanA)
1
−
cosA
1
−
cosA
1
+
(secA−tanA)
1
=0
LHS=
sec
2
A−tan
2
A
secA−tanA+secA+tanA
−
cosA
2
=2secA−
cosA
2
=
cosA
2
−
cosA
2
=0
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