If `cosx=(2cosy-1)/(2-cosy),` where `x , y in (0,pi)` then `tanx/2coty/2` is equal to `sqrt(2)` (b) `sqrt(3)` (c) `1/(sqrt(2))` (d) `1/(sqrt(3))`

Explanation

`cosx=(2cosy-1)/(2-cosy)`

`cos2theta=(1-tan^2theta)/(1+tan^2(x/2)`

`(1-tan^2(x/2))/(1+tan^2(x/2))=(2cosy-1)/(2-cosy)`

`(1-tan^2(x/2)+1+tan^2(x/2))/(1-tan^2(x/2)-1-tan^2(x/2))=(2cosy-1+2-cosy)/(2cosy-1-2+cosy)`

`2/(-2tan^2(x/2))=(wsy+1)/(3cosy-3)`

`tan^2(x/2)=(3(1-cosy))/(1+cosy)`

`tan^2(x/2)=(3(2sin^2(y/2)))/(2cos^2(y/2))`

`tan^2(x/2)=3tan^2(y/2)`

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