Prove that 1 + secθ + tanθ =
2 ÷ 1+cotθ−cosecθ
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Answer:
(1+cotθ−cosecθ)(1+tanθ+secθ)=2
L.H.S (1+cotθ−cosecθ)(1+tanθ+secθ)
=(1+
sinθ
cosθ
−
sinθ
1
)(1+
cosθ
sinθ
+
cosθ
1
)
=(
sinθ
sinθ+cosθ−1
)(
cosθ
sinθ+cosθ+1
)
=
sinθcosθ
(sinθ+cosθ)
2
−1
=
sinθcosθ
2sinθcosθ+sin
2
θ+cos
2
θ−1
=
sinθcosθ
1+2sinθcosθ−1
=
sinθcosθ
2sinθcosθ
=2
LHS=RHS
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