Class 12 Maths

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

  1. `0≤y≤π`
  2. `-(π)/(2)≤y≤(π)/(2)`
  3. `0<y<π`
  4. `-(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

  1. π
  2. `-(π)/(3)`
  3. `(π)/(3)`
  4. `(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)`



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