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)`



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