Class 12 Maths

Inverse Trignometric Functions

NCERT Solution

Miscellaneous Exercise 2 Part 1

Prove The Following:

Question 12: (9π)/8-9/4text(sin)^(-1)1/3=9/4text(sin)^(-1)(2sqrt2)/3

Solution: Given, (9π)/8-9/4text(sin)^(-1)1/3=9/4text(sin)^(-1)(2sqrt2)/3

Or, (9π)/8=9/4text(sin)^(-1)(2sqrt2)/3+9/4text(sin)^(-1)1/3

Now, RHS =9/4text(sin)^(-1)(2sqrt2)/3+9/4text(sin)^(-1)1/3

=9/4(text(sin)^(-1)(2sqrt2)/3+text(sin)^(-1)1/3)

=9/4[text(sin)^(-1)(1/3sqrt(1-((2sqrt2)/3)^2)+(2sqrt2)/3sqrt(1-(1/3)^2))]

=9/4[text(sin)^(-1)(1/3sqrt(1/9)+(2sqrt2)/3sqrt(8/9))]

=9/4[text(sin)^(-1)(1/3xx1/3+(2sqrt2)/3xx(2sqrt2)/3)]

=9/4[text(sin)^(-1)(1/9+8/9)]

=9/4(text(sin)^(-1)9/9)

=9/4(text(sin)^(-1)(1))

=9/4xx(π)/2=(9π)/8 = LHS proved

Solve the following equations:

Question 13: 2text(tan)^(-1)(text(cos) x)=text(tan)^(-1)(2text(cosec) x)

Solution: Given, 2text(tan)^(-1)(text(cos) x)=text(tan)^(-1)(2text(cosec) x)

Or, text(tan)^(-1)((2text(cos) x)/(1-text(cos)^2x)=text(tan)^(-1)((2x)/(1-x^2))

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

Or, text(tan)^(-1)((2text(cos) x)/(text(sin)^2x))=text(tan)^(-1)(2/(text(sin) x))

Or, (2text(cos) x)/(text(sin)^2x)=2/(text(sin) x)

Or, (2text(cos) x)/(text(sin)^2x)xx(text(sin) x)/2=1

Or, (text(cos) x)/(text(sin) x)=1

Or, text(cot) x=1

Or, x=text(cot)^(-1)(1)

Or, x=(π)/4

Question 14: text(tan)^(-1)(1-x)/(1+x)=1/2text(tan)^(-1)x, (x>0)

Solution: Given, text(tan)^(-1)(1-x)/(1+x)=1/2text(tan)^(-1)x

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

Or, text(tan)^(-1)[(2((1-x)/(1+x)))/(1-((1-x)/(1+x))^2)]=text(tan)^(-1)x

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

Or, text(tan)^(-1)[(2(1-x)/(1+x))/(((1+x)^2-(1-x)^2)/((1+x)^2))]=text(tan)^(-1)x

Or, text(tan)^(-1)[((2(1-x))/(1+x))/((1+x^2+2x-(1+x^2-2x))/((1+x)^2))]=text(tan)^(-1)x

Or, text(tan)^(-1)[((2(1-x))/(1+x))/((1+x^2+2x-1-x^2+2x)/((1+x)^2))]=text(tan)^(-1)x

Or, text(tan)^(-1)[((2(1-x))/(1+x))/((4x)/((1+x)^2))]=text(tan)^(-1)x

Or, text(tan)^(-1)[(2(1-x))/(1+x)xx((1+x)^2)/(4x)]=text(tan)^(-1)x

Or, text(tan)^(-1)x[((1-x)(1+x))/(2x)]=text(tan)^(-1)x

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

Or, (1-x^2)/(2x)=x

Or, 1-x^2=2x^2

Or, 2x^2+x^2=1

Or, 3x^2=1

Or, x^2=1/3

Or, x=±1/(sqrt3)

But, as per question, x>0

So, x=1/(sqrt3)

Question 15: text(sin)(text(tan)^(-1)x), |x|<1 is equal to

1. x/(sqrt(1-x^2))
2. 1/(sqrt(1-x^2))
3. 1/(sqrt(1+x^2))
4. x/(sqrt(1+x^2))

Answer: (d) x/(sqrt(1+x^2))

Explanation: Given, text(sin)(text(tan)^(-1)x)

=text(sin)(text(sin)^(-1)x/(sqrt(1+x^2)))

Because text(tan)^(-1)x=text(sin)^(-1)xx(x)/(sqrt(1+x^2))

Question 16: text(sin)^(-1)(1-x)-2text(sin)^(-1)x=(π)/2, then x is equal to

1. 0, 1/2
2. 1, 1/2
3. 0
4. 1/2

Explanation: Let, x=text(sin)θ

Or, θ=text(sin)^(-1)x

Given, text(sin)^(-1)(1-x)-2text(sin)^(-1)x=(π)/2

Or, text(sin)^(-1)(1-text(sin)θ)-2θ=(π)/2

Or, -2θ=(π)/2-text(sin)^(-1)(1-text(sin)θ)

Since, text(sin)^(-1)x+text(cos)^(-1)x=(π)/2v

Or, text(cos)^(-1)x=(π)/2-text(sin)^(-1)x

So, -2θ=text(cos)^(-1)(1-text(sin)θ)

Or, text(cos)(-2θ)=1-text(sin)θ

Or, text(cos)2θ=1-text(sin)θ

Or, 1-2text(sin)^2θ=1-text(sin)θ

Or, 1-2text(sin)^2θ=1-text(sin)θ

Or, 1-1-2(sin)^2θ=-text(sin)θ

Or, -2text(sin)^2θ=-text(sin)θ

Or, 2text(sin)^2θ=text(sin)θ

Or, 2text(sin)^2θ-text(sin)θ=0

Or, text(sin)θ(2text(sin)θ-1)=0

After substituting x=text(sin)θ

x(2x-1)=0

Hence, either x=0 or 2x-1=0

If, 2x-1=0

Then, x=1/2

After substituting this value in LHS we get;

text(sin)^(-1)(1-1/2)-2text(sin)^(-1)1/2

=text(sin)^(-1)1/2-2text(sin)^(-1)1/2

Or, -text(sin)^(-1)1/2=-(π)/6

But, -(π)/6≠(π)/2

Hence, x=1/2 does not satisfy the given expression.

So, x=0 is correct

Question 17: text(tan)^(-1)x/y-text(tan)^(-1)(x-y)/(x+y) is equal to

1. (π)/2
2. (π)/3
3. (π)/4
4. -(3π)/4

Answer: (c) (π)/4

Explanation: Given, text(tan)^(-1)x/y-text(tan)^(-1)(x-y)/(x+y)

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

So, given expression can be written as follows:

text(tan)^(-1)[(x/y-(x-y)/(x+y))/(1+x/y\xx(x-y)/(x+y))]

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

=text(tan)^(-1)[(x^2+xy-xy+y^2)/(x^2+y^2-xy+xy)]

=text(tan)^(-1)(1)=(π)/4`

Exercise 2.1

Exercise 2.2

Misc Exercise Part 1

Misc Exercise Part 2

Misc Exercise Part 3