tanA/1-cotA + cotA/1-cotA = 1+ secA × cosecA
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Answer:
+
1−tanA
cotA
=1+secAcscA
Taking L.H.S.-
1−cotA
tanA
+
1−tanA
cotA
=
1−(
tanA
1
)
tanA
+
1−tanA
(
tanA
1
)
=
tanA−1
tan
2
A
+
tanA(1−tanA)
1
=
tanA(1−tanA)
1−tan
3
A
=
tanA(1−tanA)
(1−tanA)(1+tanA+tan
2
A)
(∵a
3
−b
3
=(a−b)(a
2
+ab+b
2
))
=
tanA
sec
2
A+tanA
(∵1+tan
2
A=sec
2
A)
=1+
tanA
sec
2
A
=1+
cosAsinA
1
=1+secAcscA
= R.H.S.
Hence proved
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