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

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