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
- `(7π)/6`
- `(5π)/6`
- `(π)/3`
- `(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
- `1/2`
- `1/3`
- `1/4`
- 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
- π
- `-(π)/2`
- 0
- `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`