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Tardigrade
Question
Mathematics
If y = (2 sin α/1 + cos α + sin α) then ( 1 - cos α + sin α /1 + sin α) is equal to
Q. If
y
=
1
+
c
o
s
α
+
s
i
n
α
2
s
i
n
α
then
1
+
s
i
n
α
1
−
c
o
s
α
+
s
i
n
α
is equal to
3310
250
Trigonometric Functions
Report Error
A
1 + y
12%
B
1 - y
17%
C
y
1
21%
D
y
49%
Solution:
Now,
1
+
s
i
n
α
1
−
c
o
s
α
+
s
i
n
α
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
(
1
−
c
o
s
α
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
(
1
+
s
i
n
α
)
2
−
c
o
s
2
α
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
1
+
s
i
n
2
α
+
2
s
i
n
α
−
(
1
−
s
i
n
2
α
)
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
1
+
s
i
n
2
α
+
2
s
i
n
α
−
1
+
s
i
n
2
α
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
2
s
i
n
α
+
2
s
i
n
2
α
=
(
1
+
s
i
n
α
)
(
1
+
c
o
s
α
+
s
i
n
α
)
2
s
i
n
α
(
1
+
s
i
n
α
)
=
1
+
c
o
s
α
+
s
i
n
α
2
s
i
n
α
=
y