pls help me solve this
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Answered by
0
cos a = 3/5
cos
2
A=
(
3
5
)
2
=
9
25
Since
sin
2
A+
cos
2
A=1, so we have
sin
2
A=1−
cos
2
A
=1−
(
3
5
)
2
=1−
9
25
=
25−9
25
=
16
25
So the given expression beocmes
9
cot
2
A−1=9
cos
2
A
sin
2
A
−1
(since cotA=
cosA
sinA
)
=9
9
25
16
25
−1
=9×
9
16
−1
=
81
16
−1
=
81−16
16
=
65
16
cos
2
A=
(
3
5
)
2
=
9
25
Since
sin
2
A+
cos
2
A=1, so we have
sin
2
A=1−
cos
2
A
=1−
(
3
5
)
2
=1−
9
25
=
25−9
25
=
16
25
So the given expression beocmes
9
cot
2
A−1=9
cos
2
A
sin
2
A
−1
(since cotA=
cosA
sinA
)
=9
9
25
16
25
−1
=9×
9
16
−1
=
81
16
−1
=
81−16
16
=
65
16
Answered by
0
cos a = 3/5
cos
2
A=
(
3
5
)
2
=
9
25
Since
sin
2
A+
cos
2
A=1, so we have
sin
2
A=1−
cos
2
A
=1−
(
3
5
)
2
=1−
9
25
=
25−9
25
=
16
25
So the given expression beocmes
9
cot
2
A−1=9
cos
2
A
sin
2
A
−1
(since cotA=
cosA
sinA
)
=9
9
25
16
25
−1
=9×
9
16
−1
=
81
16
−1
=
81−16
16
=
65
16
cos
2
A=
(
3
5
)
2
=
9
25
Since
sin
2
A+
cos
2
A=1, so we have
sin
2
A=1−
cos
2
A
=1−
(
3
5
)
2
=1−
9
25
=
25−9
25
=
16
25
So the given expression beocmes
9
cot
2
A−1=9
cos
2
A
sin
2
A
−1
(since cotA=
cosA
sinA
)
=9
9
25
16
25
−1
=9×
9
16
−1
=
81
16
−1
=
81−16
16
=
65
16
rosy1746:
seriously i didn't understand anything
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