if tanA=3/4 and A is not in the first quadrant then {sin(π/2+A) - cot(π-A)} / tan(3π/2-A) - cos(3π/2+A)
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Answer:
Given
π
<
x
<
3
π
2
and
tan
x
=
3
4
π
<
x
<
3
π
2
⇒
π
2
<
x
2
<
3
π
4
→
x
2
∈
2nd quadrant
This means
sin
(
x
2
)
→
+
v
e
cos
(
x
2
)
→
−
v
e
tan
(
x
2
)
→
−
v
e
Now
tan
x
=
3
4
⇒
2
tan
(
x
2
)
1
−
tan
2
(
x
2
)
=
3
4
⇒
8
tan
(
x
2
)
=
3
−
3
tan
2
(
x
2
)
⇒
3
tan
2
(
x
2
)
+
8
tan
(
x
2
)
−
3
=
0
⇒
3
tan
2
(
x
2
)
+
9
tan(x2)−tan(x2)−3=0
⇒
3tan(x2)(tan(x2)+3)−1(tan(x2)+3)=0
⇒(3tan
(x2)−1)(tan(x2)
+3)=0
This means
tan
(x2)
=13
→
not acceptable as
tan(x2)
→−ve
So
tan(x2)
→−3
Now
cos(x2)
=1sec(x2)
=−1√1+tan2(x2)
=−1√1+(−3)2
=−1√10
Again
sin(x2)
=tan(x2)×cos(x2)
=−3×(−1√10)
=3√10
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