Class 12 Maths

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`



Exercise 2.1 Part 1

Exercise 2.1 Part 2

Exercise 2.2 Part 1

Exercise 2.2 Part 2

Exercise 2.2 Part 3

Exercise 2.2 Part 4

Miscellaneous Exercise