Is O not the origin (0,0)?
Then angle QOP should be 90 degrees or pi/2.
The posted diagram and the text of the question don't tally.
In the diagram above, Ox and Oy are the rectangular axes and the straight line PAQ passes through the fixed point A ( 4,1 ). The angle QPO= Theta and 0 < Theta < pi/2
Let T = theta ( Sorry , I do not know how to insert the symbol, I tried &Theta but no luck )
a) Show thatOP + OQ = 5 + 4 tan T + cot T and show that as T changes, the least value of OP + OQ is 9.
b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes.
Let OP = u, and OQ = v
In the right triangle below the point (4,1):
vertical leg = 1
horizontal leg = (u -4)
tanT = 1 / (u -4)
u -4 = 1 / tanT = cotT
u = cotT +4 -----------**
In the right triangle above the (4,1):
vertical leg = (v -1)
horizontal leg = 4
tanT = (v -1) / 4
4tanT = v -1
v = 4tanT +1
So, (OP +OQ) = (u +v) = cotT +4 +4tanT +1
(oP +OQ) = 5 +4tanT +cotT -----------------------shown.
For the least value of (OP +OQ), get the derivative of (OP +OQ) with respect to T. Then equate that to zero.
You should arrive to 4 sec^2(T) = csc^2(T).
Which is the same as cot^2(T) = 4
Or, cotT = 2, and tanT = 1/2
Then minimum (OP +OQ) = 9
---------------------------------------
b) Show that PQ = 4 sec T + cosec T and find, correct to two significant figures, the minimum value of Q when T changes
In the said right triangle below (4,1):
hypotense = say, r
sinT = 1/r
r = 1/sinT = cscT
In the right triangle above the (4,1):
hypotenuse = say, s
cosT = 4/s
s = 4/cosT = 4secT
Thus, PQ = r +s = cscT +4secT
PQ = 4secT +cscT ----------------------shown.
For the minimum value of Q.....?
Q? or PQ?