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