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