178k views
2 votes
A+B+C = Pi & SecB = SecC SecA Then Prove Tan A=2cot C

1 Answer

4 votes

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.

User Treddy
by
7.8k points

Related questions

asked Jan 15, 2021 36.5k views
Ould Abba asked Jan 15, 2021
by Ould Abba
8.1k points
2 answers
1 vote
36.5k views
asked Aug 23, 2021 128k views
Mtekeli asked Aug 23, 2021
by Mtekeli
7.6k points
1 answer
4 votes
128k views
asked Mar 25, 2024 125k views
Razack asked Mar 25, 2024
by Razack
8.3k points
1 answer
5 votes
125k views