If angle of projection are `((pi)/4 + theta)` and `((pi)/4 -theta)` where `theta lt(pi)/4`, then the ration of horizontal ranges described by the projectile is (speed is same)-

tan(pi/4 + theta) tan (pi/4 -theta)=

tan(pi/4 + theta) + tan(pi/4-theta)=4

tan((pi)/(4)+theta)-tan((pi)/(4)-theta)=?

Two projectiles are projected angle ((pi)/(4) + theta) and ((pi)/(4) - theta) with the horizontal , where theta lt (pi)/(4) , with same speed. The ratio of horizontal ranges described by them is

Two projectiles are projected angle ((pi)/(4) + theta) and ((pi)/(4) - theta) with the horizontal , where theta lt (pi)/(4) , with same speed. The ratio of horizontal ranges described by them is

If tan((pi)/(4)+theta)+tan((pi)/(4)-theta)=4 then theta=

Simplify: tan(pi/4 + theta) tan ((3pi)/4 + theta)

Tan ((pi)/(4)+theta).Tan((3pi)/(4)+theta)=

If tan (pi//4 + theta) + tan (pi//4 - theta) = 4 , then theta =