Find the principal value of the following (i) `cos^(-1) (cos 3)` (ii) `cos^(-1) (cos 4)` (iii) `cos^(-1) (cos 15)` (iv) `cos^(-1) (cos 30)` (v) `cos^(

(i) ` sin^(-1).(1)/(2) = sin ^(-1) (sin .(pi)/(6)) = (pi)/(6)` (ii) `tan^(-1).(1)/(sqrt(3)) = tan^(-1)(tan .(pi)/(6))=(pi)/(6)` (iii) `cot^(-1)(-sqrt(3)) = pi - cot^(-1) (sqrt(3))` ` = pi cot^(-1)(cot.(pi)/(6))` ` = pi -(pi)/(6)...

`cos^(-1)(-(1)/(2))=pi-cos^(-1)((1)/(2))=pi-cos^(-1)(cos""(pi)/(3))=pi-(pi)/(3)=(2pi)/(3)`

(i) Since `1 in [-pi//2, pi//2], sin^(-1) (sin 1) = 1` (ii) Since `2 in [-pi//2, pi//2], sin^(-1) (sin 2) != 2` `:. Sin^(-1) (sin 2) = sin^(-1) (sin (pi)...

Correct Answer - A::C Given equation is `x^(2) + 2x sin(cos^(-1) y) + 1 =0` Since x is real, `D ge 0`. Therefore, `4(sin(cos^(-1)y))^(2) -4 ge 0` or `(sin(cos^(-1)y))^(2) ge 1`...

Correct Answer - B `sin^(-1) x in [-(pi)/(2), (pi)/(2)]` `cos^(-1) y in [0, pi]` `sec^(-1) z in [0, (pi)/(2)) uu ((pi)/(2), pi]` `rArr sin^(-1) x + cos^(-1) y + sec^(-1) z...

Correct Answer - B We have `f(x) = sin^(-1)x`, having range `[(pi)/(2), (3pi)/(2)]` Consider the function `h(x) = (pi - sin^(-1) x)`, where `sin^(-1) x in [-(pi)/(2), (pi)/(2)]` Clearly, function `f(x)...

Correct Answer - i) `pi/3`, ii) `pi/6`, iii) `pi/3`, iv) `pi/4`, v) `pi/6`, vi) `pi/6`, vii) `pi/4`.

Correct Answer - i) `-pi/4`, ii) `(5pi)/6`, iii) `-pi/3`, iv) `(2pi)/3`, v) `-pi/4`, vi) `(2pi)/3`

Correct Answer - C `(c )` `arg(z_(1)z_(2))=arg(z_(1))+arg(z_(2))+2mpi`, `m in I` `:.` Principal argument `=alpha+beta-2pi`

Correct Answer - B `(b)` Clearly `alpha=-i` where `i^(2)=-1` So `Delta(alpha)=0+alpha^(2)+1-1-0-alpha^(3)` `=(-i)^(2)+1-1-(-i)^(3)` =-1+1-1-i=-1-i` So, principal argument of `Delta(alpha)` is `-(3pi)/(4)`.