Given that a particle is thrown with **angle of projection** #theta# over a triangle #DeltaACB# from one of its end #A# of the horizontal base #AB# aligned along X-axis and it finally falls at the other end #B#of the base,grazing the vertex #C(x,y)#

Let #u# be the velocity of projection, #T# be the time of flight, #R=AB# be the horizontal range and #t# be the time taken by the particle to reach at C #(x,y)#

The horizontal component of velocity of projection #->ucostheta#

The vertical component of velocity of projection #->usintheta#

Considering motion under gravity without any air resistance we can write

#y=usinthetat-1/2 g t^2.....[1]#

#x=ucosthetat...................[2]#

**combining [1] and [2] we get**

#y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/(u^2cos^2theta)#

#=>y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/u^2xxsec^2theta#

#=>color(blue)(y/x=tantheta-((gsec^2theta)/(2u^2))x........[3])#

Now during time of flight #T# **the vertical displacement is zero**

So

#0=usinthetaT-1/2 g T^2#

#=>T=(2usintheta)/g#

Hence horizontal displacement during time of flight i.e. range is given by

#R=ucosthetaxxT=ucosthetaxx(2usintheta)/g=(u^2sin2theta)/g#

#=>R=(2u^2tantheta)/(g(1+tan^2theta))#

#=>R=(2u^2tantheta)/(gsec^2theta)#

#=>color(blue)((gsec^2theta)/(2u^2)=tantheta/R......[4])#

**Combining [3] and [4] we get**

#y/x=tantheta-1/2 xx(gx)/u^2xxsec^2theta#

#=>y/x=tantheta-(xtantheta)/R#

#=>tanalpha=tantheta-(xtantheta)/R# [since #color(red)(y/x=tanalpha)# from figure]

So #tantheta=tanalphaxx(R/(R-x))#

#=>tantheta=tanalphaxx((R-x+x)/(R-x))#

#=>tantheta=tanalphaxx(1+x/(R-x))#

#=>tantheta=tanalpha+(xtanalpha)/(R-x)#

#=>tantheta=tanalpha+y/(R-x)# [putting #color(red)(xtanalpha=y)#]

Finally we have from figure #color(magenta)(y/(R-x)=tanbeta)#

Hence we get our required relation

#color(green)(tantheta=tanalpha+tanbeta)#

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