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.
$\endgroup$ 01 Answer
$\begingroup$$\dfrac{2!}{1!9!}+\dfrac{2!}{3!7!}+\dfrac{1}{5!5!}$=$\dfrac{2}{10!}$($^{10}C_1$)+$\dfrac{2}{10!}$($^{10}C_3$)+$\dfrac{1}{10!}$($^{10}C_5$)
$\implies$$\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 $\implies$ $c=7$
$\cos A+\cos B$=$\dfrac{25+49-9}{90}$+$\dfrac{9+49-25}{42}$=$\dfrac{12}{7}$.
$\endgroup$ 4
