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`



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