"A line OA of length $ r$ starts from its initial position OX and traces an angle AOB=$\alpha$ in the anticlockwise direction.It then traces back in the clockwise direction an angle BOC=3 $\theta $(where alpha is greater than 3 $\theta$).L is the foot of the perpendicular form C on OA."
" Also $\frac{\sin ^3 \theta}{CL} = \frac{\cos^3 \theta}{OL} = 1$
1) $ \frac{1-r\cos \alpha }{r \sin \alpha} $ is equal to..
2)$ \frac{2r\sin \alpha }{1+2r\cos \alpha} $ is equal to
3)$ \frac{2r^2 - 1}{r } $ is equal to ...
I ( yet) don't want help in solving the problem itself but what this gibberish mean. I did try to do it myself.
For example first sentence say "....from its initial position OX...". How can a line start from a line OX shouldn't it be point O instead??
This quote is taken from cengage JEE advanced by G.Tiwani.
I don't know if this title is good here. It will be great if you could edit that
By line $OA$ I think they mean line segment $OA.$ The symbol $OX$ can be thought of as representing a ray extending to the right. Then the situation described is a counterclockwise sweep of $OA$ about $O$ until the angle measures $\alpha.$ Then this arm stops and sweeps back for $3\theta,$ but ends up before getting back to the initial position $OA$ since we're told $3\theta<\alpha.$ This new position of the segment is called $OC.$ The perpendicular dropped from $C$ meets the axis $OX$ at $L.$ Draw a figure and consider the various right triangles involved. You might also need the triple-angle formulas.
Good luck.