Answer:
LHS=
1−cotθ
sinθ
+
1−tanθ
cosθ
=
1−(1/tanθ)
tanθ−1
sinθtanθ
sinθtanθ−cosθ
sin
2
θ
−cosθ
cosθ(tanθ−1)
θ−cos
cosθ(sinθ/cosθ−1)
(sinθ+cosθ)(sinθ−cosθ)
sinθ−cosθ)
=sinθ+cosθ
hy
Step-by-step explanation:
ela unaru brother and ur study
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Answers & Comments
Answer:
LHS=
1−cotθ
sinθ
+
1−tanθ
cosθ
=
1−(1/tanθ)
sinθ
+
1−tanθ
cosθ
=
tanθ−1
sinθtanθ
+
1−tanθ
cosθ
=
tanθ−1
sinθtanθ−cosθ
=
tanθ−1
cosθ
sin
2
θ
−cosθ
=
cosθ(tanθ−1)
sin
2
θ−cos
2
θ
=
cosθ(sinθ/cosθ−1)
(sinθ+cosθ)(sinθ−cosθ)
=
sinθ−cosθ)
(sinθ+cosθ)(sinθ−cosθ)
=sinθ+cosθ
Verified answer
Answer:
hy
Step-by-step explanation:
ela unaru brother and ur study