Trigonometric Equations Question 204
Question: In $ \Delta ABC, $ $ b^{2}\cos 2A-a^{2}\cos 2B= $
Options:
A) $ b^{2}-a^{2} $
B) $ b^{2}-c^{2} $
C) $ c^{2}-a^{2} $
D) $ a^{2}+b^{2}+c^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ b^{2}\cos 2A-a^{2}\cos 2B $ $ =b^{2}(1-2{{\sin }^{2}}A)-a^{2}(1-2{{\sin }^{2}}B) $ $ =b^{2}-a^{2}-2(b^{2}{{\sin }^{2}}A-a^{2}{{\sin }^{2}}B)=b^{2}-a^{2} $ .
BETA
