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Mathematics
Problem
Let
h
(
x
)
=
tan
(
2
x
)
cot
(
3
x
)
h(x)=\dfrac{\tan(2x)}{\cot(3x)}
h
(
x
)
=
cot
(
3
x
)
tan
(
2
x
)
. Which of the following is equivalent to
h
′
(
x
)
h'(x)
h
′
(
x
)
? a)
2
sec
2
(
2
x
)
cot
(
3
x
)
−
3
tan
(
2
x
)
csc
2
(
3
x
)
cot
2
(
3
x
)
\dfrac{2\sec^{2}(2x)\cot(3x)-3\tan(2x)\csc^{2}(3x)}{\cot^{2}(3x)}
cot
2
(
3
x
)
2
sec
2
(
2
x
)
cot
(
3
x
)
−
3
tan
(
2
x
)
csc
2
(
3
x
)
b)
2
sec
2
(
2
x
)
cot
(
3
x
)
+
3
tan
(
2
x
)
csc
2
(
3
x
)
cot
2
(
3
x
)
\dfrac{2\sec^{2}(2x)\cot(3x)+3\tan(2x)\csc^{2}(3x)}{\cot^{2}(3x)}
cot
2
(
3
x
)
2
sec
2
(
2
x
)
cot
(
3
x
)
+
3
tan
(
2
x
)
csc
2
(
3
x
)
c)
6
sec
2
(
2
x
)
csc
2
(
3
x
)
cot
2
(
3
x
)
\dfrac{6\sec^{2}(2x)\csc^{2}(3x)}{\cot^{2}(3x)}
cot
2
(
3
x
)
6
sec
2
(
2
x
)
csc
2
(
3
x
)
d)
sec
2
(
2
x
)
cot
(
3
x
)
+
3
tan
(
2
x
)
csc
2
(
3
x
)
cot
2
(
3
x
)
\dfrac{\sec^{2}(2x)\cot(3x)+3\tan(2x)\csc^{2}(3x)}{\cot^{2}(3x)}
cot
2
(
3
x
)
sec
2
(
2
x
)
cot
(
3
x
)
+
3
tan
(
2
x
)
csc
2
(
3
x
)
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