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Manipal Entrance Test 2015 Math Paper
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© examsnet.com
Question : 11
Total: 70
If
cos
−
1
x
2
+
cos
−
1
y
3
=
θ
,
then
9
x
2
−
12
x
y
cos
θ
+
4
y
2
is equal to
36
−
36
sin
2
θ
36
sin
2
θ
36
cos
2
θ
Validate
Solution:
We have,
cos
−
1
x
2
+
cos
−
1
y
3
=
θ
⇒
cos
−
1
{
x
y
6
−
√
1
−
x
2
4
.
√
1
−
y
2
9
}
=
θ
⇒
x
y
−
√
4
−
x
2
.
√
9
−
y
2
=
√
6
cos
θ
⇒
(
x
y
−
6
cos
θ
)
2
=
(
4
−
x
2
)
(
9
−
y
2
)
⇒
−
12
x
y
cos
θ
+
36
cos
2
θ
=
36
−
4
y
2
−
9
x
2
⇒
9
x
2
+
4
y
2
−
12
x
y
cos
θ
=
36
sin
2
θ
© examsnet.com
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