Trigonometric Equations Question 199
Question: In $ \Delta ABC,{{(a-b)}^{2}}{{\cos }^{2}}\frac{C}{2}+{{(a+b)}^{2}}{{\sin }^{2}}\frac{C}{2}= $
Options:
A) $ a^{2} $
B) $ b^{2} $
C) $ c^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ (a^{2}+b^{2}-2ab){{\cos }^{2}}\frac{C}{2}+(a^{2}+b^{2}+2ab){{\sin }^{2}}\frac{C}{2} $ $ =a^{2}+b^{2}+2ab( {{\sin }^{2}}\frac{C}{2}-{{\cos }^{2}}\frac{C}{2} ) $ $ =a^{2}+b^{2}-2ab\cos C=a^{2}+b^{2}-(a^{2}+b^{2}-c^{2})=c^{2} $ .
BETA
