Class 12 Maths

Inverse Trignometric Functions

NCERT Solution

Exercise 2.2 Part 1

Find the principal values of the following:

Question 1: 3text(sin)^(-1)x=text(sin)^(-1)(3x-4x^3), x∈[-1/2,1/2]

Solution: Let text(sin)^(-1)x=θ

So, x=text(sin)θ)

Now, LHS =3text(sin)^(-1)x=3θ

Again, RHS =text(sin)^(-1)(3x-4x^3)

Or, text(sin)^(-1)(3text(sin)θ-4text(sin)θ)

=text(sin)^(-1)(text(sin)3θ)=3θ

Thus, LHS = RHS

Question 2: 3text(cos)^(-1)x=text(cos)^(-1)(4x^2-3x), x∈[1/2,1]

Solution: Let text(cos)^(-1)x=θ

So, text(cos)θ=x

LHS =3text(cos)^(-1)x=3θ

RHS =text(cos)^(-1)(4x^&2-3x)

=text(cos)^(-1)(4text(cos)^3θ-3text(cos)θ)

=text(cos)^(-1)(text(cos)θ)=3θ = LHS proved

Question 3: text(tan)^(-1)(2)/(11)+text(tan)^(-1)(7)/(24)=text(tan)^(-1)1/2

Solution: LHS: text(tan)^(-1)(2)/(11)+text(tan)^(-1)(7)/(24)

=text(tan)^(-1)[((2)/(11)+(7)/(24))/(1-(2)/(11)xx(7)/(24))]

(Because text(tan)^(-1)x+text(tan)^(-1)y=(x+y)/(1-xy))

=text(tan)^(-1)[((48+77)/(264))/(1-(14)/(264))]

=text(tan)^(-1)[(124)/(250)]

=text(tan)^(-1)1/2= RHS proved

Question 4: 2text(tan)^(-1)1/2+text(tan)^(-1)1/7=text(tan)^(-1)(31)/(17)

Solution: LHS =2text(tan)^(-1)1/2+text(tan)^(-1)1/7

=text(tan)^(-1)[(2xx1/2)/(1-(1/2)^2)]+text(tan)^(-1)1/7

[Because 2text(tan)^(-1)x=text(tan)^(-1)[(2x)/(1-x^2)]]

=text(tan)^(-1)[1/(1-1/4)]+text(tan)^(-1)1/7

=text(tan)^(-1)4/3+text(tan)^(-1)1/7

=text(tan)^(-1)[(4/3+1/7)/(1-4/3xx1/7)]

[Because text(tan)^(-1)x+text(tan)^(-1)y=(x+y)/(1-xy)]

=text(tan)^(-1)[((28+3)/(21))/(1-(4)/(21))]

=text(tan)^(-1)[((31)/(21))/((21-4)/(21))]

=text(tan)^(-1)(31)/(17)= RHS proved

Write the following functions in the simplest form

Question 5: text(tan)^(-1)(sqrt(1+x^2)-1)/x, x≠0

Solution: Let, x=text(tan)θ

So, θ=text(tan)^(-1)x

Now, text(tan)^(-1)(sqrt(1+x^2)-1)/x

=text(tan)^(-1)(sqrt(1+text(tan)^2θ)-1)/(text(tan)θ)

=text(tan)^(-1)(sqrt(sec)^2θ)-1)/(text(tan)θ

=text(tan)^(-1)((text(sec)θ-1)/(text(tanθ))

=text(tan)^(-1)((text(sec)θ-1)/(text(tan)θ))

=text(tan)^(-1)(((1)/(text(cos)θ)-1)/((text(sin)θ)/(text(cos)θ)))

=text(tan)^(-1)(((1-text(cos)θ)/(text(cos)θ))/((text(sin)θ)/(text(cos)θ))

=text(tan)^(-1)(1-(text(cos)θ)/(text(sin)θ))

=text(tan)^(-1)((2text(sin)^2θ(θ)/(2))/(2text(sin)(θ)/(2)text(cos)(θ)/(2)))

=text(tan)^(-1)((text(sin)^2θ(θ)/(2))/( text(cos)(θ)/(2)))

=text(tan)^(-1)(text(tan)(θ)/(2))=(θ)/(2)

=1/2text(tan)^(-1)x

So, text(tan)^(-1)(sqrt(1+x^2)-1)/(x)=1/2text(tan)^(-1)x