tanA/(1+secA) + (1+secA)/tanA
Answers
Answered by
2
Answer:
Answer
1+secA
tanA
−
1−secA
tanA
=2⋅cosecA
L.H.S
1+secA
tanA
−
1−secA
tanA
=
1−sec
2
A
tanA(1−secA)−tanA(1+secA)
=
−tan
2
A
tanA(1−secA−1−secA)
=
−tanA
−2secA
=
sinA/cosA
2⋅1/cosA
=
sinA
2
=2⋅cosecA
=R.H.S
∴ L.H.S=R.H.S.
Answered by
0
Answer:
Answer
1+secA
tanA
−
1−secA
tanA
=2⋅cosecA
L.H.S
1+secA
tanA
−
1−secA
tanA
=
1−sec
2
A
tanA(1−secA)−tanA(1+secA)
=
−tan
2
A
tanA(1−secA−1−secA)
=
−tanA
−2secA
=
sinA/cosA
2⋅1/cosA
=
sinA
2
=2⋅cosecA
=R.H.S
∴ L.H.S=R.H.S.
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