Class 12 Maths

Inverse Trignometric Functions

NCERT Solution

Miscellaneous Exercise 2 Part 1

Prove The Following:

Question 6: `text(cos)^(-1)(12)/(13)+text(sin)^(-1)3/5``=text(sin)^(-1)(56)/(65)`

Solution: LHS `=text(cos)^(-1)(12)/(13)+text(sin)^(-1)3/5`

`=text(sin)^(-1)sqrt(1-((12)/(13))^2)+text(sin)^(-1)3/5`

`=text(sin)^(-1)sqrt(1-(144)/(169))+text(sin)^(-1)3/5`

`=text(sin)^(-1)sqrt((169-144)/(169))+text(sin)^(-1)3/5`

`=text(sin)^(-1)sqrt((25)/(169))+text(sin)^(-1)3/5`

`=text(sin)^(-1)5/(13)+text(sin)^(-1)3/5`

`=text(sin)^(-1)[5/(13)sqrt(1-(3/5)^2)+3/5sqrt(1-(5/(13))^2)]`

`=text(sin)^(-1)[5/(13)sqrt((25-9)/(25))+3/5sqrt((169-25)/(169))]`

`=text(sin)^(-1)[5/(13)xx4/5+3/5xx(12)/(13)]`

`=text(sin)^(-1)[4/(13)+(36)/(65)]`

`=text(sin)^(-1)[(20+36)/(65)]``=text(sin)^(-1)(56)/(65)` = RHS proved


Question 7: `text(tan)^(-1)(63)/(16)``=text(sin)^(-1)5/(13)+text(cos)^(-1)3/5`

Solution: RHS `=text(sin)^(-1)5/(13)+text(cos)^(-1)3/5`

`=text(sin)^(-1)5/(13)+text(sin)^(-1)sqrt(1-(3/5)^2)`

`=text(sin)^(-1)5/(13)+text(sin)^(-1)sqrt((25-9)/(25))`

`=text(sin)^(-1)5/(13)+text(sin)^(-1)4/5`

`=text(sin)^(-1)[5/(13)sqrt(1-(4/5)^2)+4/5sqrt((5/(13))^2)]`

`=text(sin)^(-1)[5/(13)sqrt((25-16)/(25))+4/5sqrt((169-25)/(169))]`

`=text(sin)^(-1)[5/(13)xx3/5+4/5xx(12)/(13)]`

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

`=text(tan)^(-1)[(15+48)/(65)]=text(sin)^(-1)(63)/(65)`

`=text(tan)^(-1)[((63)/(65))/(sqrt(1-((63)/(65))^2))]`

`=text(tan)^(-1)[((63)/(65))/(sqrt((4225-3969)/(4225)))]`

`=text(tan)^(-1)[((63)/(65))/(sqrt((256)/(4225)))]``=text(tan)[((63)/(65))/((16)/(65))]`

`=text(tan)^(-1)[(63)/(65)xx(65)/(16)]=text(tan)^(-1)(63)/(16)` = LHS proved

Question 8: `text(tan)^(-1)1/5+text(tan)^(-1)1/7+text(tan)^(-1)1/3+text(tan)^(-1)1/8``=(π)/4`

Solution: LHS `=text(tan)^(-1)1/5+text(tan)^(-1)1/7+text(tan)^(-1)1/3+text(tan)^(-1)1/8`

`=(text(tan)^(-1)1/5+text(tan)^(-1)1/7)+(text(tan)^(-1)1/3+text(tan)^(-1)1/8)`

`=text(tan)^(-1)((1/5+1/7)/(1-1/5xx1/7))+text(tan)^(-1)((1/3+1/8)/(1-1/3xx1/8))`

`=text(tan)(((7+5)/(35))/((35-1)/(35)))+text(tan)^(-1)(((8+3)/(24))/((24-1)/(24)))`

`=text(tan)^(-1)(12)/(34)+text(tan)^(-1)(11)/(23)`

`=text(tan)^(-1)(((12)/(34)+(11)/(23))/(1-(12)/(34)xx(11)/(23)))`

`=text(tan)^(-1)(((276+374)/(782))/((782-132)/(782)))`

`=text(tan)^(-1)(650)/(650)`

`=text(tan)^(-1)(1)=(π)/4` = RHS proved

Question 9: `text(tan)^(-1)sqrtx=1/2text(cos)^(-1)((1-x)/(1+x))`, `x∈[0,1]`

Solution: RHS `=1/2text(cos)^(-1)((1-x)/(1+x))`

Let, `x=text(tan)^2θ`

Hence, `sqrt(x)=text(tan)θ`

So, `θ=text(tan)^(-1)sqrt(x)`

Hence, RHS `=1/2text(cos)^(-1)((1-text(tan)^2θ)/(1+text(tan)^2θ))`

`=1/2text(cos)^(-1)(text(cos)2θ)`

`=1/2xx2θ=θ`

`=text(tan)^(-1)sqrt(x)` = LHS proved



Question 10: `text(cot)^(-1)[(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))/(sqrt(1+text(sin)x)-sqrt(1-text(sin)x))]``=x/2`, `x∈(0,(π)/4)`

Solution: LHS `=text(cot)^(-1)[(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))/(sqrt(1+text(sin)x)-sqrt(1-text(sin)x))]`

After multiplying with `(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))/(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))`

`=text(cot)^(-1)[(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))/(sqrt(1+text(sin)x)-sqrt(1-text(sin)x))``xx``(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))/(sqrt(1+text(sin)x)+sqrt(1-text(sin)x))]`

`=text(cot)^(-1)[(2+2sqrt(1-text(sin)^2x))/(2text(sin)x)]`

`=text(cot)^(-1)[(2(1+text(cos)x))/(2text(sin)x)]`

`=text(cot)^(-1)[(1+text(cos)x)/(text(sin)x)]`

`=text(cot)^(-1)[(2text(cos)^2x/2)/(2text(sin)x/2.text(cos)x/2)]`

`=text(cot)^(-1)[(text(cos)x/2)/(text(sin)x/2)]`

`=text(cot)^(-1)(text(cot)x/2)`

`=x/2` = RHS proved

Question 11: `text(tan)^(-1)[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]``=(π)/4-1/2text(cos)^(-1)x`. `-1/(sqrt2)≤x≤1`

Solution: LHS `=text(tan)^(-1)[(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]`

Let `x=text(cos)2θ`

So, `θ=1/2text(cos)^(-1)x`

So, the given expression can be written as follows:

`text(tan)^(-1)[(sqrt(1+text(cos)2θ)-sqrt(1-text(cos)2θ))/ (sqrt(1+text(cos)2θ)+sqrt(1-text(cos)2θ))]`

`=text(tan)^(-1)[(sqrt(2text(cos)^2θ)-sqrt(2text(sin)^2θ))/ (sqrt(2text(cos)^2θ)+sqrt(2text(sin)^2θ))]`

`=text(tan)^(-1)[(sqrt(2)(text(cos)θ-text(sin)θ))/ (sqrt(2)(text(cos)θ+text(sin)θ))]`

`=text(tan)^(-1)[(text(cos)θ-text(sin)θ)/ (text(cos)θ+text(sin)θ)]`

`=text(tan)^(-1)[(1-text(tan)θ)/(1+text(tan)θ)]`

`=text(tan)^(-1)[text(tan)(π)/4-θ)]`

`=(π)/4-θ`

`=(π)/4-1/2text(cos)^(-1)x` = RHS proved



Exercise 2.1

Exercise 2.2

Misc Exercise Part 1

Misc Exercise Part 2

Misc Exercise Part 3