Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
Value of ( sin 13° cos 47°+ cos 13° sin 47°/ cos 72° cos 12°+ sin 72° sin 12°) is
Q. Value of
c
o
s
7
2
∘
c
o
s
1
2
∘
+
s
i
n
7
2
∘
s
i
n
1
2
∘
s
i
n
1
3
∘
c
o
s
4
7
∘
+
c
o
s
1
3
∘
s
i
n
4
7
∘
is
125
157
Trigonometric Functions
Report Error
A
1
25%
B
0
19%
C
3
1
23%
D
3
33%
Solution:
c
o
s
7
2
∘
c
o
s
1
2
∘
+
s
i
n
7
2
∘
s
i
n
1
2
∘
s
i
n
1
3
∘
c
o
s
4
7
∘
+
c
o
s
1
3
∘
s
i
n
4
7
∘
=
c
o
s
(
7
2
∘
−
1
2
∘
)
s
i
n
(
1
3
∘
+
4
7
∘
)
=
c
o
s
6
0
∘
s
i
n
6
0
∘
=
tan
6
0
∘
=
3