Trigonometric Equations Question 198
Question: In $ \Delta ABC, $ if $ {{\sin }^{2}}\frac{A}{2},{{\sin }^{2}}\frac{B}{2},{{\sin }^{2}}\frac{C}{2} $ be in H. P. then a, b, c will be in
Options:
A) A. P.
B) G. P.
C) H. P.
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \frac{1}{{{\sin }^{2}}\frac{A}{2}},\frac{1}{{{\sin }^{2}}\frac{B}{2}},\frac{1}{{{\sin }^{2}}\frac{C}{2}} $ are in A. P.
Þ $ \frac{1}{{{\sin }^{2}}\frac{C}{2}}-\frac{1}{{{\sin }^{2}}\frac{B}{2}}=\frac{1}{{{\sin }^{2}}\frac{B}{2}}-\frac{1}{{{\sin }^{2}}\frac{A}{2}} $
Þ $ \frac{ab}{(s-a)(s-b)}-\frac{ac}{(s-a)(s-c)} $ $ =\frac{ac}{(s-a)(s-c)}-\frac{bc}{(s-b)(s-c)} $
Þ $ ( \frac{a}{s-a} ),( \frac{b(s-c)-c(s-b)}{(s-b)(s-c)} ) $ = $ ( \frac{c}{s-c} ),( \frac{a(s-b)-b(s-a)}{(s-a)(s-b)} ) $
Þ $ abs-abc-acs+abc=acs-abc-bcs+abc $
Þ $ ab-ac=ac-bc\Rightarrow ab+bc=2ac $ or $ \frac{1}{c}+\frac{1}{a}=\frac{2}{b} $ , i.e., $ a,b,c $ are in H. P. Note: Students should remember this question as a fact.
BETA
