find the value of tan^-1[(2sin^-1 root 3/2]
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Answer:
First solve the inside bracket value 2cos
−1
2
3
The range of the principal value of cos
−1
x is [0,π]
Let cos
−1
2
3
=x
cosx=
2
3
x=
6
π
where x∈[0,π]
2cos
−1
2
3
=2
6
π
=
3
π
Then the expression reduces to tan
−1
(2sin
3
π
)
Now we are going to reduce the term 2sin
3
π
We know that sin
3
π
=
2
3
2sin
3
π
=2
2
3
=
3
The expression becomes, tan
−1
(
3
)
We know that tan
−1
(
3
)=
3
π
Therefore, tan
−1
[2sin(2cos
−1
2
3
)]=
3
π
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