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Problem
Problem
A
B
C
D
ABCD
A
B
C
D
is a trapezium such that
A
B
AB
A
B
and
C
D
CD
C
D
are parallel and
B
C
⊥
C
D
BC \perp CD
B
C
⊥
C
D
, if
∠
A
D
B
=
θ
\angle ADB = \theta
∠
A
D
B
=
θ
,
B
C
=
p
BC = p
B
C
=
p
and
C
D
=
q
CD = q
C
D
=
q
, then
A
B
AB
A
B
is equal to [2013 JEE Main] a.
(
p
2
+
q
2
)
sin
θ
p
cos
θ
+
q
sin
θ
\frac{(p^2 + q^2) \sin \theta}{p \cos \theta + q \sin \theta}
p
c
o
s
θ
+
q
s
i
n
θ
(
p
2
+
q
2
)
s
i
n
θ
b.
p
2
+
q
2
cos
θ
p
cos
θ
+
q
sin
θ
\frac{p^2 + q^2 \cos \theta}{p \cos \theta + q \sin \theta}
p
c
o
s
θ
+
q
s
i
n
θ
p
2
+
q
2
c
o
s
θ
c.
p
2
+
q
2
p
2
cos
θ
+
q
2
sin
θ
\frac{p^2 + q^2}{p^2 \cos \theta + q^2 \sin \theta}
p
2
c
o
s
θ
+
q
2
s
i
n
θ
p
2
+
q
2
d.
(
p
2
+
q
2
)
sin
θ
(
p
cos
θ
+
q
sin
θ
)
2
\frac{(p^2 + q^2) \sin \theta}{(p \cos \theta + q \sin \theta)^2}
(
p
c
o
s
θ
+
q
s
i
n
θ
)
2
(
p
2
+
q
2
)
s
i
n
θ
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