​Q.NO. 77plzzzzz

Dear student,
given $\cos x=\frac{2\cos y-1}{2-\cos y}$
as we know that $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
$\Rightarrow \frac{1-{\tan }^2{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{2\left({\displaystyle \frac{1-{\tan }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)-1}{2-\left(\frac{1-{\tan }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2{\displaystyle \raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}$
let $\tan \frac{x}{2}=a and \tan \frac{y}{2}=b$.now above equation will become
$\frac{1-a^2}{1+a^2}=\frac{2\left({\displaystyle \frac{1-b^2}{1+b^2}}\right)-1}{2-\left({\displaystyle \frac{1-b^2}{1+b^2}}\right)}$
$\Rightarrow \frac{1-a^2}{1+a^2}=\frac{2(1-b^2)-(1+b^2)}{2(1+b^2)-(1-b^2)}$
$\Rightarrow \frac{1-a^2}{1+a^2}=\frac{2-2b^2-1-b^2}{2+2b^2-1+b^2}$
$\Rightarrow \frac{1-a^2}{1+a^2}=\frac{1-3b^2}{1+3b^2}$
$\Rightarrow (1-a^2)(1+3b^2)=(1+a^2)(1-3b^2)$
$\Rightarrow 2a^2=6b^2$
$\Rightarrow a^2=3b^2$
or $\frac{a^2}{b^2}=3$
or $\frac{a}{b}=\sqrt{3}$
or $\frac{\tan {\displaystyle \frac{x}{2}}}{\tan {\displaystyle \frac{y}{2}}}=\sqrt{3}$
or $\tan \left(\frac{x}{2}\right).cot\left(\frac{y}{2}\right)=\sqrt{3}$

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