Inverse Trignometric Functions
NCERT Solution
Exercise 2.1 Part 2
Find the principal values of the following:
Question 9: `text(cos)^(-1)((-1)/(sqrt2))`
Solution: Let `text(cos)^(-1)((-1)/(sqrt2))=θ`
So, `text(cos)θ=(-1)/(sqrt2)`
Or, `text(cos)θ=text(cos)(π-(π)/(4))`
Or, `text(cos)θ=text(cos)(3π)/(4)`
Since value of `text(cos)^(-1)x` lies between 0 and π. Hence, principal value of `text(cos)^(-1)((-1)/(sqrt2))` is `(3π)/(4)`
Question 10: `text(cosec)^(-1)(-sqrt2)`
Solution: Let `text(cosec)^(-1)(-sqrt2)=θ`
So, `text(cosec)^(-1)θ=-sqrt2`
Or, `text(cosec)^(-1)θ=cosec(-(π)/(4))`
Since value of `text(cosec)^(-1)x` is `[(-π)/(2), (π)/(2)]-(0)`. Hence, principal value of `text(cosec)^(-1)(-sqrt2)` is `(-π)/(4)`
Question 11: `text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2)`
Solution: Let `text(tan)^(-1)(1)=θ_1`
Or, `text(tan)θ_1=1=text(tan)(π)/(4)`
Or, `θ_1=(π)/4`
Let `text(cos)^(-1)(-1/2)=θ_2`
Or, `text(cos)θ_2=-1/2=-text(cos)((π)/(3))`
Or, `text(cos)θ_2=text(cos)(π-(π)/(3))`
Or, `text(cos)θ_2=text(cos)((2π)/(3))`
Or, `θ_2=(2π)/3`
Let `text(sin)^(-1)(-1/2)=θ_3`
Or, `text(sin)θ_3=-1/2`
Or, `text(sin)θ_3=-text(sin)((π)/(6))=text(sin)(-(π)/(6))`
Or, `θ_3=-(π)/(6)`
Now, `text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2)`
`=θ_1+θ_2+θ_3`
`=(π)/(4)+(2π)/(3)-(π)/(6)`
`=(9π)/(12)=(3π)/4`
Question 12: `text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2)`
Solution: Let `text(cos)^(-1)(1/2)=θ`
Or, `text(cos)θ=1/2=text(cos)((π)/(3))`
Or, `θ=(π)/(3)`
Let `text(sin)^(-1)(1/2)=θ_1`
Or, `text(sin)θ_1=1/2=text(sin)(π)/(6)`
Or, `θ_1=(π)/(6)`
So, `text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2)`
`θ+2θ_1`
`=(π)/(3)+(2π)/(6)`
`=(2π+2π)/(6)=(2π)/(3)`
Question 13: If `text(sin)^(-1)x=y` then
- `0≤y≤π`
- `-(π)/(2)≤y≤(π)/(2)`
- `0<y<π`
- `-(π)/(2)<y<(π)/(2)`
Answer: (b) `-(π)/(2)≤y≤(π)/(2)`
Explanation: Here, `text(sin)^(-1)x=y`
Or, `text(sin) y=x`
Since value of `text(sin)^(-1)x` lies between `-(π)/(2)` and `(π)/(2)`
Hence, `y=[-(π)/(2), (π)/(2)]`
Question 14: `text(tan)^(-1)sqrt3-text(sec)^(-1)(-2)` is equal to
- π
- `-(π)/(3)`
- `(π)/(3)`
- `(2π)/(3)`
Answer: (b) `-(π)/(3)`
Explanation: Let `text(tan)^(-1)sqrt3=θ`
Or, `text(tan)θ=sqrt3=text(tan)(π)/(3)`
Or, `θ=(π)/(3)`
Now, let `text(sec)^(-1)(-2)=θ_1`
Or, `text(sec)θ_1=text(sec)(π-(π)/(3))=text(sec)(2π)/(3)`
Or, `θ_1=(2π)/(3)`
Now, `text(tan)^(-1)sqrt3-text(sec)^(-1)(-2)=θ-θ_1`
`=(π)/(3)-(2π)/(3)=-(π)/(3)`