prove of cosec x on first principal method
please answer fast and tell me correct derivation
you have correct derivation I will mark on brainlist
Answers
Answered by
1
Answer:
As
y
=
csc
x
=
1
sin
x
,
y
+
δ
y
=
1
sin
(
x
+
δ
x
)
Hence
δ
y
=
1
sin
(
x
+
δ
x
)
−
1
sin
x
=
sin
x
−
sin
(
x
+
δ
x
)
sin
x
sin
(
x
+
δ
x
)
=
−
sin
(
x
+
δ
x
)
−
sin
x
sin
x
sin
(
x
+
δ
x
)
=
−
2
sin
(
x
+
δ
x
−
x
2
)
cos
(
x
+
δ
x
+
x
2
)
sin
x
sin
(
x
+
δ
x
)
=
−
2
sin
(
δ
x
2
)
cos
(
x
+
δ
x
2
)
sin
x
sin
(
x
+
δ
x
)
and
δ
y
δ
x
=
−
2
sin
(
δ
x
2
)
cos
(
x
+
δ
x
2
)
δ
x
sin
x
sin
(
x
+
δ
x
)
Hence
d
y
d
x
=
L
t
δ
x
→
0
δ
y
δ
x
=
L
t
δ
x
→
0
−
2
sin
(
δ
x
2
)
cos
(
x
+
δ
x
2
)
δ
x
sin
x
sin
(
x
+
δ
x
)
=
L
t
δ
x
→
0
⎛
⎜
⎜
⎝
−
sin
(
δ
x
2
)
δ
x
2
⎞
⎟
⎟
⎠
×
L
t
δ
x
→
0
⎡
⎢
⎢
⎣
cos
(
x
+
δ
x
2
)
δ
x
sin
x
sin
(
x
+
δ
x
)
⎤
⎥
⎥
⎦
=
−
1
×
cos
x
sin
2
x
=
−
1
sin
x
×
cos
x
sin
x
=
−
csc
x
cot
x
Similar questions