Trigonometric Equations Question 244
Question: If the sides of a triangle are in A. P., then the cotangent of its half the angles will be in
[MP PET 1993]
Options:
A) H. P.
B) G. P.
C) A. P.
D) No particular order
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ \cot \frac{A}{2},\cot \frac{B}{2} $ and $ \cot \frac{C}{2} $ be in A.P. , then $ 2\cot \frac{B}{2}=\cot \frac{C}{2}+\cot \frac{A}{2} $
Þ $ 2\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}} $ $ +\sqrt{\frac{s(s-a)}{(s-b)(s-c)}} $ R.H.S = $ \sqrt{\frac{s}{(s-b)}}( \sqrt{\frac{(s-c)}{(s-a)}}+\sqrt{\frac{(s-a)}{(s-c)}} ) $ $ =\sqrt{\frac{s}{s-b}}( \frac{s-c+s-a}{\sqrt{(s-a)(s-c)}} ) $ $ =\sqrt{\frac{s}{s-b}}( \frac{2s-a-c}{\sqrt{(s-a)(s-c)}} ) $ $ =2\sqrt{\frac{s}{(s-b)}}\sqrt{\frac{{{(s-b)}^{2}}}{(s-a)(s-c)}} $ , $ {\because a+c=2b $ , $ a+b+c=2s $ i.e., $ 2(s-b)=2s-a-c} $ $ =2\sqrt{\frac{s(s-b)}{(s-a)(s-c)}} $ =L.H.S. Note: Students should remember this question as a fact.
BETA
