Class 12 Maths

Inverse Trignometric Functions

NCERT Solution

Exercise 2.2 Part 2

Prove the following:

Question 6: `text(tan)^(-1)1/(sqrt(x^2-1))`, `|x|>1`

Solution: Let `x=text(sec)θ`

So, `θ=text(sec)^(-1)x`

Given, `text(tan)^(-1)1/(sqrt(x^2-1))`

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

`=text(tan)^(-1)[1/(sqrt(text(tan)^2θ))]`

`=text(tan)^(-1)[1/(text(tan)θ)]`

`=text(tan)^(-1)[text(cot)θ]`

`=text(tan)^(-1)[text(tan)((π)/(2)-θ)]`

`=(π)/(2)-θ`

`=(π)/(2)-text(sec)^(-1)x`

Hence, `text(tan)^(-1)1/(sqrt(x^2-1))=(π)/(2)-text(sec)^(-1)x`


Question 7: `text(tan)^(-1)(sqrt((1-text(cos)x)/(1+text(cos)x)))`, `x<π`

Solution: Given, `text(tan)^(-1)(sqrt((1-text(cos)x)/(1+text(cos)x)))`

`=text(tan)^(-1)(sqrt((2text(sin)^2\x/2)/(2text(cos)^2\x/2)))`

[Because `text(cos)x=2text(cos)^2\x/2-1=1-2text(sin)^2\x/2`]

`=text(tan)^(-1)(sqrt(text(tan)^2\x/2))`

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

Hence, `text(tan)^(-1)(sqrt((1-text(cos)x)/(1+text(cos)x)))=x/2`

Question 8: `text(tan)^(-1)((text(cos)x-text(sin)x)/(text(cos)x+text(sin)x))`, `0<x<π`

Solution: Given, `text(tan)^(-1)((text(cos)x-text(sin)x)/(text(cos)x+text(sin)x))`

After dividing numerator and denominator by `text(cos) x`, we get

`text(tan)^(-1)(((text(cos)x-text(sin)x)/(text(cos)x))/ ((text(cos)x+text(sin)x)/(text(cos)x)))`

`=text(tan)^(-1)(((text(cos)x)/(text(cos)x)-(text(sin)x)/(text(cos)x))/((text(cos)x)/(text(cos)x)+(text(sin)x)/(text(cos)x)))`

`=text(tan)^(-1)((1-text(tan)x)/(1+text(tan)x))`

Since, `text(tan)(π)/(4)=1`, hence above expression can be written as:

`text(tan)^(-1)((text(tan)(π)/(4)-text(tan)x)/(1+text(tan)x(text(tan)(π)/(4))))`

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

Hence, `text(tan)^(-1)[text(tan)((π)/(4)-x)]=(π)/(4)-x`

Thus, `text(tan)^(-1)((text(cos)x-text(sin)x)/(text(cos)x+text(sin)x))=(π)/(4)-x`



Question 9: `text(tan)^(-1)\x/(sqrt(a^2-x^2))`, `|x|<a`

Solution: Let `x=a text(sin)θ`

Or, `x/a=text(sin)θ`

Or, `θ=text(sin)^(-1)\x/a`

Given, `text(tan)^(-1)\x/(sqrt(a^2-x^2))`

`=text(tan)^(-1)[(a text(sin)θ)/(sqrt(a^2-(a text(sin)θ)^2))]`

`=text(tan)^(-1)[(a text(sin)θ)/(sqrt(a^2-a^2text(sin)^2θ))]`

`=text(tan)^(-1)[(a text(sin)θ)/(sqrt(a^2(1- text(sin)^2θ)))]`

`=text(tan)^(-1)[(a text(sin)θ)/(sqrt(a^2text(cos)^2θ))]`

`=text(tan)^(-1)[(a text(sin)θ)/(a text(cos)θ)]`

`=text(tan)^(-1)(text(tan)θ)=θ`

`=text(sin)^(-1)x/a`

Hence, `text(tan)^(-1)\x/(sqrt(a^2-x^2))=text(sin)^(-1)x/a`

Question 10: `text(tan)^(-1)[(3a^2x-x^3)/(a^3-3ax^2)]`, `a>0`; `-a/(sqrt3)≤x≤a/(sqrt3)`

Solution: Let `x=a text(tan)θ`

Or, `θ=text(tan)^(-1)x/a`

Given, `text(tan)^(-1)[(3a^2x-x^3)/(a^3-3ax^2)]`

`=text(tan)^(-1)[(3a^2.a text(tan)θ-(a text(tan)θ)^3)/(a^3-3a(a text(tan)θ)^2)]`

`=text(tan)^(-1)[(3a^3text(tan)θ-a^3text(tan)^3θ)/(a^3-3a^3text(tan)^2θ)]`

`=text(tan)^(-1)[(a^3(3text(tan)θ-text(tan)^3θ))/(a^3(1-3text(tan)^2θ))]`

`=text(tan)^(-1)[(3text(tan)θ-text(tan)^3θ)/(1-3text(tan)^2θ)]`

`=text(tan)^(-1)(text(tan)3 θ)`

[Because `text(tan)3 θ=(3 text(tan)θ-text(tan)^3θ)/(1-3 text(tan)^2θ)`]

`=3θ=3 text(tan)^(-1)x/a`

Hence, `text(tan)^(-1)[(3a^2x-x^3)/(a^3-3ax^2)]=3 text(tan)^(-1)x/a`



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