Question

The sides a, b, c (taken in order) of a triangle $ \Delta ABC $ are in A.P. If $ \cos \alpha =\dfrac{a}{b+c} $ , $ \cos \beta =\dfrac{b}{c+a} $ , $ \cos \gamma =\dfrac{c}{a+b} $ , then $ {{\tan }^{2}}\dfrac{\alpha }{2}+{{\tan }^{2}}\dfrac{\gamma }{2} $ is equal to:
[Note: All symbols used have usual meanings in triangle ABC.]
A. 1
B. $ \dfrac{1}{2} $
C. $ \dfrac{1}{3} $
D. $ \dfrac{2}{3} $

Answer 1

Hint: Arithmetic Progression (A.P.): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.
If three numbers a, b and c are in A.P., then:
  $ b-a=c-b $
⇒ $ 2b=a+c $
Use the fact that a, b, and c are in A.P. to eliminate one of these variables in the other expressions.
Use the identity $ \cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } $ to find the values of $ {{\tan }^{2}}\dfrac{\alpha }{2} $ and $ {{\tan }^{2}}\dfrac{\gamma }{2} $ from the given expressions of $ \cos \alpha $ and $ \cos \gamma $ .
Use componendo-dividendo: If $ \dfrac{a+b}{a-b}=\dfrac{x}{y} $ , then $ \dfrac{a}{b}=\dfrac{x+y}{x-y} $ .

Complete step-by-step answer:
Since a, b and c are in A.P., we have $ 2b=a+c $ and $ c=2b-a $ .
It is given that $ \cos \alpha =\dfrac{a}{b+c} $ .
Using the half-angle formula $ \cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } $ and the given fact that $ c=2b-a $ , we get:
⇒ $ \dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}}=\dfrac{a}{3b-a} $
Using componendo-dividendo, we get:
⇒ $ \dfrac{2{{\tan }^{2}}\dfrac{\alpha }{2}}{2}=\dfrac{(3b-a)-a}{(3b-a)+a} $
⇒ $ {{\tan }^{2}}\dfrac{\alpha }{2}=\dfrac{3b-2a}{3b} $ ... (1)
Also, given that $ \cos \gamma =\dfrac{c}{a+b} $ .
⇒ $ \dfrac{1-{{\tan }^{2}}\dfrac{\gamma }{2}}{1+{{\tan }^{2}}\dfrac{\gamma }{2}}=\dfrac{2b-a}{a+b} $
Using componendo-dividendo, we get:
⇒ $ \dfrac{2{{\tan }^{2}}\tfrac{\gamma }{2}}{2}=\dfrac{(a+b)-(2b-a)}{(a+b)+(2b-a)} $
⇒ $ {{\tan }^{2}}\dfrac{\gamma }{2}=\dfrac{2a-b}{3b} $ ... (2)
Now, $ {{\tan }^{2}}\dfrac{\alpha }{2}+{{\tan }^{2}}\dfrac{\gamma }{2} $
= $ \dfrac{3b-2a}{3b}+\dfrac{2a-b}{3b} $ ... [Using (1) and (2)]
= $ \dfrac{2b}{3b} $
= $ \dfrac{2}{3} $
So, the correct answer is “Option D”.

Note: Tangent half-angle formula:
  $ \sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } $
  $ \cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } $
  $ \tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } $
In any triangle $ \Delta ABC $ , the convention is to name $ \angle A=\alpha $ , $ \angle B=\beta $ and $ \angle C=\gamma $ . The sides opposite to these angles are named a, b and c respectively.