Inverse Trignometric Functions

NCERT Solution

Miscellaneous Exercise 2 Part 1

Find the values of the following:

Question 1: `text(cos)^(-1)(text(cos)(13π)/6)`

Solution: Given, `text(cos)^(-1)(text(cos)(13π)/6)`

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

`=text(cos)^(-1)(text(cos)(π)/6)`

Since, `(π)/6∈[0,π]`

Hence, `text(cos)^(-1)(text(cos)(13π)/6)=(π)/6`

Question 2: `text(tan)^(-1)(text(tan)(7π)/6)`

Solution: Since `(7π)/6∈(-(π)/2, (π)/2)`

Hence, `text(tan)^(-1)(text(tan)(7π)/6)``=text(tan)^(-1)[text(tan)(π+(π)/6)]`

`=text(tan)^(-1)(text(tan)(π)/6)`

Now, as `(π)/6∈(-(π)/2, (π)/2)`

Hence, `text(tan)^(-1)(text(tan)(7π)/6)=(π)/6`

Question 3: Prove that `2text(sin)^(-1)3/5=text(tan)^(-1)(24)/7`

Solution: LHS `=2text(sin)^(-1)3/5`

`=text(sin)^(-1)(2xx3/5sqrt(1-(3/5)^2)`

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

`=text(sin)^(-1)(6/5sqrt(1-9/(25)))`

`=text(sin)^(-1)(6/5sqrt((16)/(25)))`

`=text(sin)^(-1)(6/5xx4/5)=text(sin)^(-1)(24)/(25)`

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

Hence, `text(sin)^(-1)(24)/(25)=text(tan)^(-1)((24)/(25))/(sqrt(1-((24)/(25))^2))`

`=text(tan)^(-1)(((24)/(25))/(sqrt(1-(576)/(625))))`

`=text(tan)^(-1)(((24)/(25))/(sqrt(625-576)/(625)))`

`=text(tan)^(-1)(((24)/(25))/(sqrt((49)/(625))))`

`=text(tan)^(-1)((24)/(25)xx(25)/7)`

`=text(tan)^(-1)(24)/7`=RHS proved

Question 4: `text(sin)^(-1)8/(17)+text(sin)^(-1)3/5=text(tan)^(-1)(77)/(36)`

Solution: LHS `= text(sin)^(-1)8/(17)+text(sin)^(-1)3/5`

`=text(sin)^(-1)[8/(17)sqrt(1-(3/5)^2)+3/5sqrt(1-(8/(17))^2)]`

`=text(sin)^(-1)[8/(17)sqrt((25-9)/(25))+3/5sqrt((289-64)/(289))`

`=text(sin)^(-1)[8/(17)sqrt((16)/(25))+3/5sqrt((225)/(289))]`

`text(sin)^(-1)[8/(17)xx4/5+3/5xx(15)/(17)]`

`=text(sin)^(-1)[(32)/(85)+(45)/(85)=text(sin)^(-1)(77)/(85)`

`=text(tan)^(-1)[((77)/(85))/(sqrt(1-((77)/(85))^2)]`

Because `text(sin)^(-1)x=text(tan)^(-1)x/(sqrt(1-x^2)`

`=text(tan)^(-1)[((77)/(85))/(sqrt((7225-5925)/(7225)))]`

`=text(tan)^(-1)[((77)/(85))/(sqrt((126)/(7225)))]`

`=text(tan)^(-1)[((77)/(85))/((36)/(85))]`

`=text(tan)^(-1)(77)/(36)`= RHS proved

Question 5: `text(cos)^(-1)4/5+text(cos)^(-1)(12)/(13)=text(cos)^(-1)(33)/(65)`

Solution: LHS `=text(cos)^(-1)4/5+text(cos)^(-1)(12)/(13)`

Since `text(cos)^(-1)x+text(cos0^(-1)y``=text(cos)^(-1)[xy-sqrt(1-x^2)sqrt(1-y^2)]`

Hence, LHS can be written as follows:

`text(cos)^(-1)[4/5xx(12)/(13)-sqrt(1-(4/5)^2)sqrt(1-((12)/(13))^2)]`

`=text(cos)^(-1)[(48)/(65)-sqrt(1-(16)/(25))sqrt(1-(144)/(169))]`

`=text(cos)^(-1)(48)/(65)-3/5xx5/(13)]`

`=text(cos)^(-1)[(48)/(65)-3/(13)]`

`=text(cos)^(-1)[(48-15)/(65)]``=text(cos)^(-1)(33)/(65)`= RHS proved