Inverse Trignometric Functions
NCERT Solution
Exercise 2.1 Part 1
Find the principal values of the following:
Question 1: `text(sin)^(-1)(-1/2)`
Solution: Let `text(sin)^(-1)(-1/2)=θ`
Therefore, `text(sin)θ=-1/2`
Or, `text(sin)θ=-text(sin)30°`
Or, `text(sin)θ=-text(sin)((π)/(6))`
Or, `text(sin)θ=text(sin)((-π)/(6))`
Since principal value of `text(sin)^(-1)x` lies between `-(π)/(2)` and `(π)/(2)`
Hence, principal value of `text(sin)^(-1)(-1/2)` is `-(π)/(6)`
Question 2: `text(cos)^(-1)((sqrt3)/(2))`
Solution: Let `text(cos)^(-1)((sqrt3)/(2))=θ`
Hence, `text(cos)θ=(sqrt3)/2`
Or, `text(cos)θ=text(cos)((π)/(6))`
Since principal value of `text(cos)^(-1)` lies between 0 and π. Hence, principal value of `text(cos)^(-1)((sqrt3)/(2))` is `(π)/6`
Question 3: `text(cosec)^(-1)(2)`
Solution: Let `text(cosec)^(-1)(2)=θ`
So, `text(cosec)θ=2`
Or, `text(cosec)θ=text(cosec)((π)/(6))`
Since principal value of `text(cosec)^(-1)x` is `[(-π)/(2), (π)/(2)]-(0)``
Hence, principal value of `text(cosec)^(-1)(2)` is `(π)/6`
Question 4: `text(tan)^(-1)(-sqrt3)`
Solution: Let `text(tan)^(-1)(-sqrt3)=θ`
So, `text(tan)θ=-sqrt3`
Or, `text(tan)θ=-text(tan)((π)/(3))`
Or, `text(tan)θ=text(tan)((-π)/(3))`
Since value of `text(tan)^(-1)x` lies between `(-π)/2` and `(π)/2`
Hence, principal value of `text(tan)^(-1)(-sqrt3)` is `(-π)/3`
Question 5: `text(cos)^(-1)(-1/2)`
Solution: Let `text(cos)^(-1)(-1/2)=θ`
So, `text(cos)θ=-1/2`
Or, `text(cos)θ=-text(cos)((π)/(3))`
Or, `text(cos)θ=text(cos)(π-(π)/(3))`
Or, `text(cos)θ=text(cos)((2π)/(3))`
We know that the range of the principal value of `text(cos)^(-1)` is `[0, x]` and `text(cos)((2π)/(3))=-1/2`
Hence, principal value of `text(cos)^(-1)(-1/2)` is `(2π)/3`
Question 6: `text(tan)^(-1)(-1)`
Solution: Let `text(tan)^(-1)(-1)=θ`
So, `text(tan)θ=-1`
Or, `text(tan)θ=-text(tan)((π)/(4))`
Or, `text(tan)θ=text(tan)(-(π)/(4))`
Since value of `text(tan)^(-1)x` lies between `-(π)/2` and `(π)/2`
Hence, principal value of `text(tan)^(-1)(-1)` is `(-π)/4`
Question 7: `text(sec)^(-1)((2)/(sqrt3))`
Solution: Let `text(sec)^(-1)((2)/(sqrt3))=θ`
So, `text(sec)θ=2/(sqrt3)`
Or, `text(sec)θ=text(sec)(π)/6`
Hence, principal value of `text(sec)^(-1)((2)/(sqrt3))` is `(π)/6`
Question 8: `text(cot)^(-1)(sqrt3)`
Solution: Let `text(cot)^(-1)(sqrt3)=θ`
So, `text(cot)θ=sqrt3`
Or, `text(cot)θ=text(cot)(π)/6`
Thus, principal value of `text(cot)^(-1)(sqrt3)` is `(π)/6`