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`