If 3 cotA= 6 sec B = -2√10
Where π/2 < A< π and
π < B < 3π/2
Evaluate cosecA - tanB
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Step-by-step explanation:
We know:
sin
2
x
+
cos
2
x
=
1
, therefore,
sin
x
=
√
1
−
cos
2
x
sin
θ
=
±
√
1
−
(
8
13
)
2
=
±
√
1
−
64
169
=
±
√
169
−
64
169
sin
θ
=
±
√
105
169
=
±
√
105
13
But because
θ
is between
3
π
2
and
2
π
, i.e in the third quadrant,
sin
θ
=
−
√
105
13
sin
2
θ
=
2
sin
θ
cos
θ
=
2
(
√
105
13
)
(
8
13
)
=
16
√
105
169
cos
2
θ
=
1
−
2
sin
2
θ
=
1
−
2
(
105
169
)
=
169
−
210
169
=
−
41
169
tan
2
θ
=
sin
2
θ
cos
2
θ
=
16
√
105
169
−
41
169
=
−
16
√
105
41
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