The sides a, b, c (taken in order) of a ΔABC are in A.P. If \(\rm \co

1.	The sides a, b, c (taken in order) of a ΔABC are in A.P. If \(\rm \cos \alpha =\frac{a}{b+c}\), \(\rm \cos \beta =\frac{b}{c+a}\), \(\rm \cos \gamma =\frac{c}{a+b}\), then \(\rm \tan^{2}\frac{\alpha }{2}+\tan^{2}\dfrac{\gamma }{2}\) is equal to:[Note: All symbols used have usual meanings in triangle ΔABC.]1. 12. \(\rm \frac12\)3. \(\rm \frac23\)4. \(\rm \frac13\)
Answer» 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)] = \(\rm \frac{2b}{3b}\) = \(\frac{2}{3}\)

The sides a, b, c (taken in order) of a ΔABC are in A.P. If \(\rm \cos \alpha =\frac{a}{b+c}\), \(\rm \cos \beta =\frac{b}{c+a}\), \(\rm \cos \gamma =\frac{c}{a+b}\), then \(\rm \tan^{2}\frac{\alpha }{2}+\tan^{2}\dfrac{\gamma }{2}\) is equal to:[Note: All symbols used have usual meanings in triangle ΔABC.]1. 12. \(\rm \frac12\)3. \(\rm \frac23\)4. \(\rm \frac13\)

Answer» 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)]

= \(\rm \frac{2b}{3b}\)

= \(\frac{2}{3}\)