Trigonometric Equations Question 242
Question: In $ \Delta ,ABC $ , $ (b^{2}-c^{2})\cot A+(c^{2}-a^{2})\cot B+(a^{2}-b^{2})\cot C= $
Options:
A) 0
B) $ a^{2}+b^{2}+c^{2} $
C) $ 2,(a^{2}+b^{2}+c^{2}) $
D) $ \frac{1}{2abc} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ (b^{2}-c^{2})\cot A=(b^{2}-c^{2})\frac{\cos A}{\sin A}=\frac{(b^{2}-c^{2})(b^{2}+c^{2}-a^{2})}{2bc.ka} $ Hence L.H.S. $ =\frac{1}{2kabc} $ $ [(b^{4}-c^{4})+(c^{4}-a^{4})+(a^{4}-b^{4})-{a^{2}(b^{2}-c^{2}) $ $ +b^{2}(c^{2}-a^{2})+c^{2}(a^{2}-b^{2})}]=0 $ .
BETA
