1+sinteta-costeta/1+sinteta+costeta
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Answer:
Consider the L.H.S
sinθ+cosθ−1
sinθ−cosθ+1
=(
sinθ+cosθ−1
sinθ−cosθ+1
)×(
sinθ+cosθ+1
sinθ+cosθ+1
)
=(
sinθ+cosθ−1
sinθ+1−cosθ
)×(
sinθ+cosθ+1
sinθ+1+cosθ
)
=
(sinθ+cosθ)
2
−1
2
(sinθ+1)
2
−cos
2
θ
=
sin
2
θ+cos
2
θ+2sinθcosθ−1
sin
2
θ+1+2sinθ−cos
2
θ
Since, sin
2
θ+cos
2
θ=1
Therefore,
=
1+2sinθcosθ−1
1−cos
2
θ+1+2sinθ−cos
2
θ
=
2sinθcosθ
2−2cos
2
θ+2sinθ
=
sinθcosθ
1−cos
2
θ+sinθ
=
sinθcosθ
sin
2
θ+sinθ
=
cosθ
sinθ+1
=
cosθ
1
+
cosθ
sinθ
=secθ+tanθ
=(secθ+tanθ)×(
secθ−tanθ
secθ−tanθ
)
=
secθ−tanθ
sec
2
θ−tan
2
θ
We know that
sec
2
θ−tan
2
θ=1
Therefore,
=
secθ−tanθ
1
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