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Byju's Answer
Standard XII
Mathematics
Trigonometric Ratios of Compound Angles
Find the valu...
Question
Find the values of each of the following:
(i)
tan
-
1
2
cos
2
sin
-
1
1
2
(ii)
cot
tan
-
1
a
+
cot
-
1
a
(iii)
cos
sec
-
1
x
+
cosec
-
1
x
,
x
≥
1
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Solution
(i) Let
sin
-
1
1
2
=
y
Then,
sin
y
=
1
2
∴
tan
-
1
2
cos
2
sin
-
1
1
2
=
tan
-
1
2
cos
2
y
=
tan
-
1
2
1
-
2
sin
2
y
∵
cos
2
x
=
1
-
2
sin
2
x
=
tan
-
1
2
1
-
2
×
1
4
∵
sin
y
=
1
2
=
tan
-
1
2
×
1
2
=
tan
-
1
1
=
π
4
∴
tan
-
1
2
cos
2
sin
-
1
1
2
=
π
4
(ii)
We
have
cot
tan
-
1
a
+
cot
-
1
a
=
cot
π
2
tan
-
1
x
+
cot
-
1
x
=
π
2
=
0
∴
cot
tan
-
1
+
cot
-
1
a
=
0
(iii)
We have
cos
sec
-
1
x
+
cosec
-
1
x
=
cos
π
2
∵
sec
-
1
x
+
cosec
-
1
x
=
π
2
=
0
∴
cos
sec
-
1
x
+
cosec
-
1
x
=
0
,
|
x
|
≥
1
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0
Similar questions
Q.
Find the values of each of the following:
(i)
tan
-
1
2
cos
2
sin
-
1
1
2
(ii)
cos
sec
-
1
x
+
cosec
-
1
x
,
x
≥
1
Q.
Evaluate:
(i)
cot
sin
-
1
3
4
+
sec
-
1
4
3
(ii)
sin
tan
-
1
x
+
tan
-
1
1
x
for
x
<
0
(iii)
sin
tan
-
1
x
+
tan
-
1
1
x
for
x
>
0
(iv)
cot
tan
-
1
a
+
cot
-
1
a
(v)
cos
sec
-
1
x
+
cosec
-
1
x
,
x
≥
1
Q.
Evaluate each of the following:
(i)
cos
-
1
1
2
+
2
sin
-
1
1
2
(ii)
tan
-
1
2
cos
2
sin
-
1
1
2
(iii)
tan
-
1
1
+
cos
-
1
-
1
2
+
sin
-
1
-
1
2
(iv)
tan
-
1
3
-
sec
-
1
(
-
2
)
+
cosec
-
1
2
3
Q.
If
cos
(
2
sin
−
1
x
)
=
1
9
, then find the value of
x
.
Q.
Inverse circular functions,Principal values of
s
i
n
−
1
x
,
c
o
s
−
1
x
,
t
a
n
−
1
x
.
t
a
n
−
1
x
+
t
a
n
−
1
y
=
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
<
1
π
+
t
a
n
−
1
x
+
y
1
−
x
y
,
x
y
>
1
.
Solve
(a)
c
o
s
(
2
s
i
n
−
1
x
)
=
1
/
9
(b)
c
o
s
−
1
(
3
/
5
)
−
s
i
n
−
1
(
4
/
5
)
=
c
o
s
−
1
x
(c) If
s
i
n
(
s
i
n
−
1
1
5
+
c
o
s
−
1
x
)
=
1
, then prove that x is equal to
1
/
5
.
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