Answer Posted / ezzuddin

The correct answer and below is the proof of the formula:

KVAR= KW X [Sqrt(1–(CosQ)^2)]/CosQ;

The proof:
Pythagoras Theorem states that in a right-angled triangle
the square of the hypotenuse is equal to the sum of the
squares of the other two sides.
so that:
C is the hypotenuse.
S is the opposite.
P is the adjacent.
TanQ= opposite/adjacent = S/P -----> S=P x(TanQ;)
Pf=CosQ;= adjacent/hypotenuse = P/C ---->C=P/CosQ;
Pythagoras said that:
C^2= P^2 + S^2
P^2/(CosQ)^2= P^2 + (P^2 x (TanQ)^2)
by divide P^2:
1/(CosQ)^2 = 1 + (TanQ)^2
by changing the both sides:
(TanQ)^2 = (1/(CosQ)^2) - 1
(TanQ)^2 = (1/(CosQ)^2) - ((CosQ)^2/(CosQ)^2)
by Taking the (1/(COSQ)^2) Common factor:
(TanQ)^2 = {1/(CosQ)^2)} x ( 1 - (CosQ)^2)
by taking the square root:

TanQ; = [Sqrt(1–(CosQ)^2)]/CosQ;

so that:
KVAR = KW x [Sqrt(1–(CosQ)^2)]/CosQ;

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