Here the situation is shown below,

So,let after time #t# of its motion,it will reach height #a#,so considering vertical motion,

we can say,

#a=(u sin theta)t -1/2 g t^2# (#u# is the projection velocity of projectile)

Solving this we get,

#t=(2u sin theta_-^+sqrt(4u^2 sin^2 theta -8ga))/(2g)#

So,one value (smaller one) of #t=t# (let) is suggesting the time to reach #a# while going up and the other (larger one) #t=t'# (let) while coming down.

So,we can say in this time interval the projectilw horizontally travelled distance #2a#,

So,we can write, #2a=u cos theta(t'-t)#

Putting the values and arranging,we get,

#u^4 sin^2 2theta -8gau^2 cos^2 theta-4a^2g^2=0#

Solving for #u^2#,we get,

#u^2 =(8gacos^2 theta _-^+sqrt(64g^2a^2 cos^4 theta+16a^2g^2sin^2 2theta))/(2 sin^2 2theta)#

Putting back #sin 2theta=2 sin theta cos theta# we get,

#u^2=(8gacos^2 theta _-^+sqrt(64g^2a^2 cos^4 theta+64a^2g^2sin^2 theta cos^2 theta))/(2 sin^2 2theta)#

or, #u^2=(8ga cos^2 theta+sqrt(64g^2a^2cos^2theta(cos^2 theta+sin^2 theta)))/(2sin^2 2theta)=(8gacos^2theta+8ag cos theta)/(2 sin^2 2theta)=(8agcostheta(cos theta+1))/(2 sin^2 2theta)#

now,formula for range of projectile motion is #R=(u^2 sin 2 theta)/g#

So,multiplying the obtained value of #u^2# with #(sin2 theta)/g#,we get,

#R=(2a(cos theta+1))/sin theta=(2a* 2 cos^2(theta/2))/(2 sin (theta/2) cos (theta/2))=2a cot (theta/2)#