Correct Answer – Option 3 : \(\rm \frac23\)

Concept:

Trigonometry: \(\rm \cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\).

Calculation:

Since a, b and c are in A.P., we have 2b = a + c ⇒ c = 2b – a.

It is given that \(\rm \cos \alpha =\frac{a}{b+c}\).

Using the half-angle formula \(\rm \cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\) and the fact that c = 2b – a, we get:

⇒ \(\rm \frac{1-{{\tan }^{2}}\frac{\alpha }{2}}{1+{{\tan }^{2}}\frac{\alpha }{2}}=\frac{a}{3b-a}\)

⇒ \(\rm \frac{2{{\tan }^{2}}\tfrac{\alpha }{2}}{2}=\frac{(3b-a)-a}{(3b-a)+a}\)

⇒ \(\rm {{\tan }^{2}}\frac{\alpha }{2}=\frac{3b-2a}{3b}\)            …(1)

Similarly, \(\rm {{\tan }^{2}}\frac{\gamma }{2}=\frac{2a-b}{3b}\)            …(2)

Now, \(\rm {{\tan }^{2}}\frac{\alpha }{2}+{{\tan }^{2}}\frac{\gamma }{2}\)

= \(\rm \frac{3b-2a}{3b}+\frac{2a-b}{3b}\)            … [Using (1) and (2)]

= \(\rm \frac{2b}{3b}\)

= \(\frac{2}{3}\)