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.