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SSC CHSL 10 Jan 2017 Shift 2 Paper
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© examsiri.com
Question :
81 of 100
Marks:
+1
,
-0
What is the value of tan3A?
(
3
t
a
n
A
−
t
a
n
3
A
)
∕
(
1
−
3
t
a
n
2
A
)
(
3
t
a
n
A
+
t
a
n
3
A
)
∕
(
1
+
3
t
a
n
2
A
)
(
3
t
a
n
A
−
t
a
n
3
A
)
∕
(
1
+
3
t
a
n
2
A
)
(
3
t
a
n
A
+
t
a
n
3
A
)
∕
(
1
−
3
t
a
n
2
A
)
Validate
Solution:
Expression :
t
a
n
(
3
A
)
=
s
i
n
(
3
A
)
c
o
s
(
3
A
)
=
s
i
n
(
2
A
+
A
)
c
o
s
(
2
A
+
A
)
=
s
i
n
(
2
A
)
c
o
s
A
+
c
o
s
(
2
A
)
s
i
n
A
c
o
s
(
2
A
)
c
o
s
A
−
s
i
n
(
2
A
)
s
i
n
A
Using,
s
i
n
(
2
x
)
=
2
s
i
n
x
c
o
s
x
and
c
o
s
(
2
x
)
=
c
o
s
2
x
−
s
i
n
2
x
=
2
s
i
n
A
c
o
s
2
A
+
s
i
n
A
(
c
o
s
2
A
−
s
i
n
2
A
)
c
o
s
A
(
c
o
s
2
A
−
s
i
n
2
A
)
−
2
s
i
n
2
A
c
o
s
A
Taking common from numerator and from denominator,
=
s
i
n
A
2
c
o
s
2
A
+
c
o
s
2
A
−
s
i
n
2
A
c
o
s
A
c
o
s
2
A
−
s
i
n
2
A
−
2
s
i
n
2
A
=
t
a
n
A
3
c
o
s
2
A
−
s
i
n
2
A
c
o
s
2
A
−
3
s
i
n
2
A
Dividing both numerator and denominator by
c
o
s
2
A
, we get :
=
t
a
n
A
3
−
t
a
n
2
A
1
−
3
t
a
n
2
A
=
3
t
a
n
A
−
t
a
n
3
A
1
−
3
t
a
n
2
A
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