Answer
See the result below.
Work Step by Step
We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$
$\dfrac{v^2}{9} [\sin 2 (45-k)^{\circ}]= \dfrac{v^2}{9} [\sin 2 (45+k)^{\circ}]$
or, $\dfrac{v^2}{9} \sin (90-2k)^{\circ}= \dfrac{v^2}{9} \sin (90+2k)^{\circ}$
or, $\dfrac{v^2}{9} \cos (2k)^{\circ}= \dfrac{v^2}{9} \cos (2k)^{\circ}$
Hence, the horizontal distance when $\theta= (45-k)^{\circ} $ with distance when $\theta= (45+k)^{\circ} $ is the same.
