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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.

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

$\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 $\implies$ $c=7$

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

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

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