Question

Let $\theta$ be angle of projection , $\alpha$ be angle of elevation to highest point (from point of projection) and $\beta$ be angle of elevation to highest point (from point where projectile meets the ground).

PROVE:
$\boxed{ \tan( \alpha ) + \tan( \beta ) = \tan( \theta) }$
​

Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}$

Let $\sf{\theta}$ be angle of projection , $\sf{\alpha}$ be angle of elevation to highest point (from point of projection) and $\sf{\beta}$ be angle of elevation to highest point (from point where projectile meets the ground).  Prove $\sf{\tan (\alpha)+\tan (\beta)=\tan (\theta)}$

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• Angle of projection = $\sf{\theta}$

• Angle of elevation to highest point (from point of projection) = $\sf{\alpha}$

• Angle of elevation to highest point (from point where projectile meets the ground) = $\sf{\beta}$

$\underline{\underline{\sf{\maltese\:\:From\:the\:Edited \:Picture}}}$

• Maximum Height of Projectile = H

• Range of Projectile = R

$\underline{\underline{\sf{\maltese\:\:To\:Prove}}}$

• $\sf{\tan (\alpha)+\tan (\beta)=\tan (\theta)}$

$\underline{\underline{\sf{\maltese\:\:Formula\:Used}}}$

• If a Projectile is fired with initial speed $\sf{u}$ at angle $\sf{\theta}$ from horizontal, then

$\boxed{\sf{H=\dfrac{u^{2} \sin ^{2} \theta}{2 g}}\:\:\:\:\:and\:\:\:\:\:R=\dfrac{u^{2} \sin2\theta}{g}}$

• Also, From Trigonometric Functions

$\boxed{\sf{\tan\theta=\dfrac{\textsf{ Perpandicular }}{\textsf{ Base}}}}$

$\boxed{\sf{\sin 2 \theta=2 \sin \theta \cos \theta}}$

$\boxed{\sf{\tan\theta=\dfrac{\sf{\sin\theta}}{\sf{\cos\theta}}}}$

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

In Triangle OAC,

$\to\sf{\tan\alpha=\dfrac{H}{R/2}=\dfrac{2H}{R}}$

In Triangle ACB,

$\to\sf{\tan\beta =\dfrac{H}{R/2}=\dfrac{2H}{R}}$

From This Values,

$\sf{\tan\alpha+\tan\beta=\dfrac{2H}{R}+\dfrac{2H}{R}=\dfrac{4H}{R}}$

Now Putting the Values of H and R, We get :

$\to\sf{\tan\alpha+\tan\beta=\dfrac{4\times H}{R}}$

$\to\sf{\tan\alpha+\tan\beta=\dfrac{4\times \dfrac{u^{2} \sin ^{2} \theta}{2 g}}{\dfrac{u^{2} \sin2\theta}{g}}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{4\times \frac{u^2\sin ^2\left(\theta\right)}{2g}g}{u^2\sin \left(2\theta\right)}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{2u^2\sin ^2\left(\theta\right)}{u^2\sin \left(2\theta\right)}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{2\sin ^2\left(\theta\right)}{\sin \left(2\theta\right)}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{2\sin ^2\left(\theta\right)}{2\sin \theta\cos\theta}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{2\sin\left(\theta\right)}{2\cos\theta}}$

$\to\sf{\displaystyle\tan\alpha+\tan\beta=\frac{\sin\theta}{\cos\theta}}$

$\boxed{\boxed{\to\sf{\displaystyle\tan\alpha+\tan\beta=\tan\theta}}}$

Hence Proved !!

$\underline{\underline{\sf{\maltese\:\:Alternate\:Method}}}$

(Kindly Refer to Attachments - 2, 3, 4 and 5 for Alternate method)

Answer 2

$\bold{Answer\;:-}$

• Given :-

• To prove : tan α + tan β = tan θ

• Solution :-

tan α = y/x

tan β = y/(R-x)

Therefore, tan α + tan β = y/x + y/R-x

• Equation of trajectory ( in the form of range ) :-

y = x*tan θ [1- (x/R)]

y/x = tan θ [(R-x)/R]

y*R/x(R-x) = tan θ - (i)

Now, tan α + tan β = y/x + y/(R-x)

tan α + tan β = y [ (R-x+x)/x(R-x)]

tan α + tan β = y*R/x(R-x) - (ii)

Now, from equation (i)

tan α + tan β = tan θ

Hence proved!

Extra :-

• Eqn. of trajectory : [ Refer to the attachment! ]

• Projectile Motion :-

• A particle which is thrown into the space and which moves under the effect of gravity only is called a projectile.

• Projectile motion is a combination of two motions : Horizontal motion and Vertical motion.

• The acceleration due to gravity causes the ball to slow down as it goes upwards.

• There is no force in the horizontal direction so there is no acceleration.

• Trajectory of a projectile is parabolic only when the acceleration of the projectile is constant and • The direction of acceleration is different from the direction of projectile is initial velocity.

its initial position and initial velocity. Under the same acceleration due to gravity, the trajectory of an object can be a straight line or a parabola depending on the initial conditions.

$\bold{Hope\;it \; helps\;!}$