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