A vertical pole subtends an angle `tan^-1 (1/2)` at apoint P on the ground. The angle subtended by the upper half of the pole at the point P is (A) `tan^-1(1/4)` (B) `tan^-1(2/9)` (C) `tan^-1(1/8)` (D) `tan^-1(2/3)`

A verrical pole subtends an angle tan^(-1)"1/2 at a points P on the ground. The angle subtended by the upper half the pole at P is

A vertical pole subtends an angle Tan^-1(1/2) at a point P on the ground. IF the angles subtended by the upper half and the lower half of the pole at P are respectively alpha and beta , then (tan alpha , tan beta) =

Show that 2 tan^-1(1/4) + 2 tan^-1 (2/9) = tan^-1(4/3) .

tan^(- 1)(1/4)+tan^(- 1)(2/9)=1/2tan^(- 1)(4/3)

tan^(- 1)(1/4)+tan^(- 1)(2/9)=1/2tan^(- 1)(4/3)

tan^(-1)((1)/(4))+tan^(-1)((2)/(9))=tan^(-1)((1)/(2))

tan^(-1)(1)/(4)+tan^(-1)(2)/(9)=tan^(-1)(1)/(2)

A vertical pole subtends an angle Tan ^(-1) (( 1)/(2)) at a point P on the ground. On the ground .If the angles subtends by the upper half and the lower half of the pole at P are respectively alpha and beta then ( tan alpha, tan beta)

The acute angle between the curve y^2=x and x^2=y at (1,1) is a) tan^-1(4/5) b) tan^-1(3/4) c) tan^-1(1) d) tan^-1(4/3)

In the given diagram shown, for a projectile, what is the angle of projection? (A) tan^(-1) (1) (B) tan^(-1)(8/3) (C) tan^(-1)(4/3) (D) tan^(-1)(5/3)