In $\triangle ABC$ , find the value of $\cos A+\cos B$

The sides of $\triangle$ABC are in Arithmetic Progression (order being $a$, $b$, $c$) and satisfy

$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}=\dfrac{8^a}{(2b)!}$, Then prove that the value of $\cos A+\cos B$ is $\dfrac{12}{7}$.

It took me a long time to solve this question, if you have another way of solving, it is appreciated.

$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}$=$\dfrac{2}{10!}$($^{10}C_1$)+$\dfrac{2}{10!}$($^{10}C_3$)+$\dfrac{1}{10!}$($^{10}C_5$)

$\implies$$\dfrac{1}{10!}(2\cdot$$^{10}C_1$+2$\cdot$$^{10}C_3+^{10}C_5) \implies$$\dfrac{1}{10!}$(2$\cdot$10+2$\cdot$120+252)=$\dfrac{512}{10!}$=$\dfrac{2^9}{10!}$

$\implies$$\dfrac{2^9}{10!}$$\implies$$\dfrac{8^3}{(2\cdot5)!}=\dfrac{8^a}{(2\cdot b)!}(given) \implies$$a=3, b=5$

If a,b,c are in AP $\implies$ $c=7$

$\cos A+\cos B$=$\dfrac{25+49-9}{90}$+$\dfrac{9+49-25}{42}$=$\dfrac{12}{7}$.

• Don't we need to prove the uniqueness of the solution? – lab bhattacharjee Dec 20 '14 at 16:17
• $2b! = (2b)!$?? – Unit Dec 20 '14 at 21:29
• yes @unit I meant $(2b)!$ – Dheeraj Kumar Dec 21 '14 at 5:54
• @labbhattacharjee Sir please explain what do you mean by uniqueness of the solution I haven't learnt that concept yet. – Dheeraj Kumar Jan 1 '15 at 9:27