Class 12 Maths

# Inverse Trignometric Functions

## NCERT Solution

### Exercise 2.2 Part 3

#### Find the values of each of the following:

Question 11: text(tan)^(-1)[2text(cos)θ(2text(sin)^(-1)1/2)]

Solution: Let text(sin)^(-1)1/2=θ

Or, text(sin)θ=1/2=text(sin)(π)/(6)

Or, text(sin)^(-1)1/2=(π)/6

Given text(tan)^(-1)[2text(cos)θ(2text(sin)^(-1)1/2)]

=text(tan)^(-1)[2text(cos)(2xx(π)/6)]

=text(tan)^(-1)[2text(cos)((π)/(3))]

=text(tan)^(-1)(2xx1/2)

=text(tan)^(-1)(1)=text(tan)^(-1)(text(tan)(π)/(4))=(π)/4

Hence, text(tan)^(-1)[2text(cos)θ(2text(sin)^(-1)1/2)] =(π)/4

Question 12: text(cot)[text(tan)^(-1)a+text(cot)^(-1)a]

Solution: Since, text(tan)^(-1)x+text(cot)^(-1)x=(π)/2, x∈R

So, the given expression can written as follows:

text(cot)((π)/2)=0

Alternate Method:

Since text(cot)^(-1)x=text(tan)^(-1)1/x

So, text(cot)[text(tan)^(-1)a+text(cot)^(-1)a]=text(cot)[text(tan)^(-1)a+text(tan)^(-1)a]

=text(cot)[text(tan)^(-1)(a+1/a)/(1-a\xx\1/a)]

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

=text(cot)[text(tan)^(-1)(a+1/a)/(0)]=text(cot)((π)/2)=0

Question 13: text(tan)1/2{text(sin)^(-1)(2x)/(1+x^2)+text(cos)^(-1)(1-y^2)/(1+y^2)}, |x|<1, y>0 and xy<1

Solution: Since text(sin)^(-1)(2x)/(1+x^2)=2text(tan)^(-1)x and text(cos)^(-1)(1-y^2)/(1+y^2)=2text(tan)^(-1)y

Hence, given expression can be written as follows:

text(tan)1/2(2text(tan)^(-1)x+2text(tan)^(-1)y)

=text(tan)1/2xx2(text(tan)^(-1)x+text(tan)^(-1)y)

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

Hence, text(cot)[text(tan)^(-1)a+text(cot)^(-1)a] =(x+y)/(1+xy)

Question 14: If text(sin)(text(sin)^(-1)1/5+text(cos)^(-1)x)=1, then find the value of x.

Solution: Using text(cos)^(-1)x=text(sin)^(-1)sqrt(1-x^2) given expression can be written as follows:

text(sin)(text(sin)^(-1)1/5+text(sin)^(-1)sqrt(1-x^2))

Or, text(sin)[text(sin)^(-1)(1/5sqrt(1-1+x^2)+sqrt(1-x^2)sqrt(1-1/(25))]=1

Because text(sin)^(-1)x+text(sin)^(-1)y=text(sin)^(-1)(x\sqrt(1-y^2)+y\sqrt(1-x^2))

Or, text(sin)[text(sin)^(-1)(1/5sqrt(s^2)+sqrt(1-x^2)sqrt((24)/(25))]=1

Or, 1/5x+sqrt(1-x^2)sqrt((24)/(25))=1

=sqrt(1-x^2)sqrt((24)/(25))=1-x/5

After squaring both sides, we get

(1-x^2)xx(24)/(25)=1-x/5

Or, (24-24x^2)/(25)=1+(x^2/(25)-(2x)/5)

Or, (24-24x^2)/(25)=(24+x^2-10x)/(25)

Or, 24-24x^2=24+x^2-10x

Or, 25+x^2-10-24+24x^2=0

Or, 25x^2-10x+1=0

Or, 25x^2-5x-5x+1=0

Or, 5x(5x-1)-1(5x-1)=0

Or, (5x-1)(5x-1)=0

Or, 5x-1=0

Or, 5x=1

Or, x=1/5

Question 15: If text(tan)^(-1)(x-1)/(x-2)+text(tan)^(-1)(x+1)/(x+2)=(π)/4, then find the value of x

Solution: text(tan)^(-1)(x-1)/(x-2)+text(tan)^(-1)(x+1)/(x+2)=(π)/4

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

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

Or, text(tan)^(-1){((x^2+2x-x-2)+(x^2-2x+x-2))/((x^2-4)-(x^2-1))}=(π)/4

Or, text(tan)^(-1){(x^2+2x-x-2+x^2-2x+x-2)/(x^2-4+x^2+1)}=(π)/4

Or, text(tan)^(-1)[(2x^2-4)/(-3)]=(π)/4

Or, (2x^2-4)/(-3)=text(tan)(π)/4=1

Or, 2x^2-4=-3

Or, 2x^2=-3+4=1

Or, x^2=1/2

Or, x=±1/(sqrt2)