Final answer:
To prove tan A = 2cot C with the conditions A + B + C = Pi and SecB = SecC SecA, we use trigonometric identities and algebraic manipulations involving A, B, C to reach the required equality.
Step-by-step explanation:
The problem at hand requires us to prove that tan A = 2cot C given that A + B + C = Pi and SecB = SecC SecA. We know from trigonometric identities that tan A = sin A / cos A and cot C = cos C / sin C. Using the given equation A + B + C = Pi, we can derive that A = Pi - B - C.
Now, using the trigonometric identity sin(Pi - x) = sin x and cos(Pi - x) = -cos x, we get sin A = sin(Pi - B - C) = sin(B + C), which simplifies further using sin(x + y) = sin x cos y + cos x sin y. For cos A, we use the fact that cos(Pi - x) = -cos x.
Since SecB = 1 / cos B and SecC = 1 / cos C, and we're given SecB = SecC SecA, we can use this relationship to find a connection between cos B and cos C. This linkage allows us to express A, B, C in terms of trigonometric functions, and with further algebraic manipulations, we can arrive at tan A = 2cot C, thereby proving the given statement.