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Byju's Answer
Standard XII
Mathematics
Evaluation of a Determinant
Evaluate the ...
Question
Evaluate the following:
(i)
sin
-
1
sin
5
π
6
(ii)
cos
-
1
cos
-
π
4
(iii)
tan
-
1
tan
3
π
4
(iv)
sin
-
1
(
sin
2
)
(v)
sin
π
3
-
sin
-
1
-
3
2
(vi)
cos
cos
-
1
-
3
2
+
π
4
(vii)
cos
tan
-
1
3
4
(viii)
cos
-
1
cos
5
π
4
(ix)
cos
-
1
cos
4
π
3
(x)
tan
-
1
tan
2
π
3
(xi)
cos
-
1
cos
13
π
6
(xii)
tan
-
1
tan
7
π
6
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Solution
We know
sin
θ
=
θ
if
-
π
2
≤
θ
≤
π
2
cos
θ
=
θ
if
0
≤
θ
≤
π
tan
θ
=
θ
i
f
-
π
2
<
θ
<
π
2
(i) We have
sin
-
1
sin
5
π
6
=
sin
-
1
sin
π
-
π
6
=
sin
-
1
sin
π
6
=
π
6
(ii) We have
cos
-
1
cos
-
π
4
=
cos
-
1
cos
π
4
=
π
4
(iii) We have
tan
-
1
tan
3
π
4
=
tan
-
1
tan
π
-
π
4
=
tan
-
1
-
tan
π
4
=
tan
-
1
tan
-
π
4
=
-
π
4
(iv) We have
sin
-
1
sin
2
=
sin
-
1
sin
π
-
2
∵
2
∉
-
π
2
,
π
2
=
π
-
2
(v) We have
sin
π
3
-
sin
-
1
-
3
2
=
sin
π
3
-
si
n
-
1
sin
-
π
3
∵
-
π
3
∈
-
π
2
,
π
2
=
sin
π
3
-
-
π
3
=
sin
2
π
3
=
3
2
∴
sin
π
3
-
sin
-
1
-
3
2
=
3
2
(vi) We have
cos
cos
-
1
-
3
2
+
π
4
=
cos
cos
-
1
cos
5
π
6
+
π
4
=
cos
5
π
6
+
π
4
=
cos
13
π
12
=
cos
π
+
π
12
=
-
cos
π
12
=
-
cos
π
4
-
π
6
=
-
cos
π
4
cos
π
6
+
sin
π
4
sin
π
6
=
-
1
2
×
3
2
+
1
2
×
1
2
=
-
3
+
1
2
2
(vii) We have
cos
tan
-
1
3
4
=
cos
1
2
cos
-
1
1
-
3
4
2
1
+
3
4
2
∵
2
tan
-
1
x
=
cos
-
1
1
-
x
2
1
+
x
2
=
cos
1
2
cos
-
1
7
25
Let
y
=
cos
-
1
7
25
⇒
cos
y
=
7
25
Now,
cos
1
2
cos
-
1
7
25
=
cos
1
2
y
=
cos
y
+
1
2
∵
cos
2
x
=
2
cos
2
x
-
1
=
7
25
+
1
2
=
32
50
=
4
5
∴
cos
tan
-
1
3
4
=
4
5
(viii) We have
cos
-
1
cos
5
π
4
=
cos
-
1
cos
2
π
-
3
π
4
=
cos
-
1
cos
3
π
4
=
3
π
4
(ix) We have
cos
-
1
cos
4
π
3
=
cos
-
1
cos
2
π
-
2
π
3
=
cos
-
1
cos
2
π
3
=
2
π
3
(x) We have
tan
-
1
tan
2
π
3
=
tan
-
1
tan
π
-
π
3
=
tan
-
1
-
tan
π
3
=
tan
-
1
tan
-
π
3
=
-
π
3
(xi) We have
cos
-
1
cos
13
π
6
=
cos
-
1
cos
2
π
+
π
6
=
cos
-
1
cos
π
6
=
π
6
(xii) We have
tan
-
1
tan
7
π
6
=
tan
-
1
tan
π
+
π
6
=
tan
-
1
tan
π
6
=
π
6
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0
Similar questions
Q.
Evaluate the following:
i.
sin
−
1
(
2
π
4
)
ii.
cos
−
1
(
cos
7
π
6
)
iii.
tan
−
1
(
tan
2
π
3
)
iv.
cos
(
cos
−
1
(
√
3
2
)
+
π
6
)
Q.
Evaluate the following:
i.
sin
−
1
(
sin
π
4
)
ii.
cos
−
1
(
cos
2
π
3
)
ii
i.
tan
−
1
(
tan
π
3
)
Q.
Evaluate the following:
(i)
tan
2
tan
-
1
1
5
-
π
4
(ii)
tan
1
2
cos
-
1
5
3
(iii)
sin
1
2
cos
-
1
4
5
(iv)
sin
2
tan
-
1
2
3
+
cos
tan
-
1
3
Q.
Evaluate each of the following:
i
tan
-
1
1
+
cos
-
1
-
1
2
+
sin
-
1
-
1
2
(ii)
tan
-
1
-
1
3
+
tan
-
1
-
3
+
tan
-
1
sin
-
π
2
(iii)
tan
-
1
tan
5
π
6
+
cos
-
1
cos
13
π
6
Q.
For the principal values, evaluate the following:
(i)
sin
-
1
1
2
-
2
sin
-
1
1
2
(ii)
sin
-
1
-
1
2
+
2
cos
-
1
-
3
2
(iii)
tan
-
1
(
-
1
)
+
cos
-
1
-
1
2
(iv)
sin
-
1
-
3
2
+
cos
-
1
3
2
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