Class 12 Maths

Inverse Trignometric Functions

NCERT Solution

Exercise 2.2 Part 4

Find the values of each of these expressions:

Question 16: `text(sin)^(-1)(text(sin)(2π)/3)`

Solution: Given, `text(sin)^(-1)(text(sin)(2π)/3)`

`=text(sin)^(-1)text(sin)(π-(π)/3)`

`=text(sin^(-1)text(sin)(π)/3``=(π)/3`

Since `(π)/3∈[-(π)/3, (π)/3]`

Hence, `text(sin)^(-1)(text(sin)(2π)/3)=(π)/3`


Question 17: `text(tan)^(-1)(text(tan)(3π)/4)`

Solution: Given, `text(tan)^(-1)(text(tan)(3π)/4)`

`=text(tan)^(-1)[text(tan)(π-(&pil)/4)]`

`=text(tan)^(-1)(-text(tan)(π)/4)`

`=-(π)/4`

Question 18: `text(tan)(text(sin)^(-1)3/5+text(cot)^(-1)3/2)`

Solution: Given, `text(tan)(text(sin)^(-1)3/5+text(cot)^(-1)3/2)`

Since `text(sin)^(-1)x=text(tan)^(-1)x/(sqrt(1-x^2))` and `text(cot)^(-1)x=text(tan)^(-1)1/x`

Hence, given expression can be written as follows:

`text(tan)(text(tan)^(-1)(3/5)/(sqrt(1-(3/5)^2))+text(tan)^(-1)1/(3/2))`

`=text(tan)(text(tan)(3/5)/(sqrt(1-9/(25)))+text(tan)^(-1)2/3)`

`=text(tan)(text(tan)^(-1)(3/5)/(sqrt((16)/(25)))+text(tan)^(-1)2/3)`

`=text(tan)(text(tan)^(-1)3/5xx5/4+text(tan)^(-1)2/3)`

`=text(tan)(text(tan)^(-1)3/4+text(tan)^(-1)2/3)`

`=text(tan)[text(tan)^(-1)(3/4+2/3)/(1-3/4xx2/3)]`

`=text(tan)[text(tan)^(-1)((17)/(12))/(1-6/(12))]`

`=text(tan)[text(tan)^(-1)(17)/(12)xx(12)/(6)]`

`=text(tan)[text(tan)^(-1)(17)/6]=(17)/6`

Hence, `text(tan)(text(sin)^(-1)3/5+text(cot)^(-1)3/2)=(17)/6`

Question 19: `text(cos)^(-1)(text(cos)(7π)/6)` is equal to

  1. `(7π)/6`
  2. `(5π)/6`
  3. `(π)/3`
  4. `(π)/6`

Answer: (b) `(5π)/6`

Explanation: Given, `text(cos)^(-1)(text(cos)(&7π)/6)`

`=text(cos)^(-1)[text(cos)(2π(5π)/6)]`

Since, `(5π)/6∈[0, π]`

Hence, `text(cos)^(-1)(text(cos)(&7π)/6)=(5π)/6`



Question 20: `text(sin)[(π)/3-text(sin)^(-1)(-1/2)]` is equal to

  1. `1/2`
  2. `1/3`
  3. `1/4`
  4. 1

Answer: (d) 1

Explanation: Given, `text(sin)[(π)/3-text(sin)^(-1)(-1/2)]`

`=text(sin[(π)/3+text(sin)^(-1)(1/2)]`

`=text(sin)[(π)/3+(π)/6]`

`=text(sin)[(3π)/6]=text(sin)[(π)/2]=1`

Question 21: `text(tan)^(-1)sqrt3-text(cot)^(-1)(-sqrt3)` is equal to

  1. π
  2. `-(π)/2`
  3. 0
  4. `sqrt3`

Answer: (b) `-(π)/2`

Explanation: Here, `text(tan)^(-1)sqrt3-text(cot)^(-1)(-sqrt3)`

`=text(tan)^(-1)sqrt3-(π-text(cot)^(-1)sqrt3)`

[Because `text(cot)^(-1)(-x)=x-text(cot)^(-1)x`]

`=text(tan)^(-1)sqrt3-π+text(cot)^(-1)sqrt3`

`=text(tan)^(-1)sqrt3-π+text(tan)^(-1)1/(sqrt3)`

`=[text(tan)^(-1)(sqrt3+1/(sqrt3))/(1-sqrt3xx1/(sqrt3)]-π`

`=text(tan)^(-1)[text(tan)^(-1)(sqrt3+1/(sqrt3))/0]-π`

`=text(tan)^(-1)∞-π`

`=(π)/2-π=-(π)2`



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