Trigonometric Equations Question 147
Question: If in a $ \Delta ABC $ , $ \cos A+2\cos B+\cos C=2 $ , then $ a,b,c $ are in
Options:
A) A. P.
B) H. P.
C) G. P.
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \cos A+2\cos B+\cos C=2 $
Þ $ \cos A+\cos C=2(1-\cos B) $
Þ $ 2\cos \frac{A+C}{2}\cos \frac{A-C}{2}=4{{\sin }^{2}}\frac{B}{2} $
Þ $ 2\cos ( \frac{A-C}{2} )=4\sin \frac{B}{2} $
Þ $ 2\cos \frac{B}{2}\cos ( \frac{A-C}{2} )=2( 2\sin \frac{B}{2}\cos \frac{B}{2} ) $
Þ $ 2\sin ,( \frac{A+C}{2} )\cos ( \frac{A-C}{2} )=2( 2\sin \frac{B}{2}\cos \frac{B}{2} ) $
Þ $ \sin A+\sin C=2\sin B\Rightarrow a+c=2b $
Þ a, b, c are in A.P.
BETA
