Prove that tan theta + sec theta - 1 / tan theta - sec theta + 1 = 1+ sin theta / cos theta
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To Prove:
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Solution:
L.H.S =
tanθ−secθ+1
tanθ+secθ−1
We can write, sec
2
θ−tan
2
θ=1
=
tanθ−secθ+1
tanθ+secθ−(sec
2
θ−tan
2
θ)
=
tanθ−secθ+1
tanθ+secθ−(secθ−tanθ)(secθ+tanθ)
=
tanθ−secθ+1
(tanθ+secθ){1−(secθ−tanθ)}
=
tanθ−secθ+1
(tanθ+secθ){1−secθ+tanθ}
=tanθ+secθ
=
cosθ
sinθ
+
cosθ
1
=
cosθ
1+sinθ
= R.H.S
since L.H.S = R.H.S
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Hence Proved.
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Solution:
L.H.S =
tanθ−secθ+1
tanθ+secθ−1
We can write, sec
2
θ−tan
2
θ=1
=
tanθ−secθ+1
tanθ+secθ−(sec
2
θ−tan
2
θ)
=
tanθ−secθ+1
tanθ+secθ−(secθ−tanθ)(secθ+tanθ)
=
tanθ−secθ+1
(tanθ+secθ){1−(secθ−tanθ)}
=
tanθ−secθ+1
(tanθ+secθ){1−secθ+tanθ}
=tanθ+secθ
=
cosθ
sinθ
+
cosθ
1
=
cosθ
1+sinθ
= R.H.S
since L.H.S = R.H.S
tanθ−secθ+1
tanθ+secθ−1
=
cosθ
1+sinθ
Hence Proved.
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