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BITSAT 2006 Solved Question Paper
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© examsnet.com
Question : 15
Total: 150
If
12
cot
2
θ
−
31
cosec
θ
+
32
=
0
, then the value of
sin
θ
is :
3
5
or 1
2
3
or
−
2
3
4
5
or
3
4
±
1
2
Validate
Solution:
12
cot
2
θ
−
31
cosec
θ
+
32
=
0
⇒
12
cos
2
θ
−
31
sin
θ
+
32
sin
2
θ
=
0
⇒
12
(
1
−
sin
2
θ
)
−
31
sin
θ
+
32
sin
2
θ
=
0
⇒
20
sin
2
θ
−
31
sin
θ
+
12
=
0
This is a quadratic equation in
sin
θ
.
∴
sin
θ
=
31
±
√
31
2
−
4
.20
.12
2
.20
=
31
±
√
961
−
960
40
=
31
±
1
40
⇒
sin
θ
=
4
5
,
3
4
© examsnet.com
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