If `sinx=(2t)/(1+t^2)` , `tany=(2t)/(1-t^2)` , find `(dy)/(dx)` .

Answer

N/a

Explanation

`:' sin x = (2t)/(1+t^(2))"....."(i)`
and ` tany = (2t)/(1-t^(2)) "...."(ii)`
`:.(d)/(dx) sin x . (dx)/(dt) = (d)/(dt)((2t)/(1+t^(2)))`
`rArr cosx (dx)/(dt) = ((1+t^(2)).(d)/(dt)(2t)-(2t).(d)/(dt)(1+t^(2)))/((1+t^(2))^(2))`
`= (2(1+t^(2))-2t.2t)/((1+t^(2))^(2)) = (2+2t^(2)-4t^(2))/((1+t^(2))^(2))`
`rArr (dx)/(dt) = (2(1-t^(2)))/((1+t^(2))^(2)).(1)/(cosx)`
`rArr (dx)/(dt) = (2(1-t^(2)))/((1+t^(2))^(2)).(1)/(sqrt(1-sin^(2)x))= (2(1-t^(2)))/((1+t^(2))^(2)).(1)/(sqrt(1-((2t)/(1+t^(2)))^(2)))`
`rArr (dx)/(dt) = (2(1-t^(2)))/((1+t^(2))^(2)).((1+t^(2)))/((1-t^(2)) ) = (2)/(1+t^(2))"......"(iii)`
Also, `(d)/(dy)tany.(dy)/(dt)=(d)/(dt)((2t)/(1-t^(2)))`
`sec^(2)y'(dy)/(dt)((1-t^(2))'(d)/(dt).(2t)-2t.(d)/(dt)(1-t^(2)))/((1-t^(2))^(2))`
`(dy)/(dt) = (2-2t^(2)+4t^(2))/((1-t^(2))^(2)).(1)/(sec^(2)y)`
`= (2(1+t^(2)))/((1-t^(2))^(2)).(1)/((1+tan^(2)y))=(2(1+t^(2)))/((1-t^(2))^(2)).(1)/(1+(4t^(2))/((1-t^(2))^(2)))`
`= (2(1+t^(2)))/((1-t^(2))^(2)).((1-t^(2))^(2))/((1+t^(2))^(2))=(2)/(1+t^(2))"....."(iv)`
`:. (dy)/(dx) = (dy//dt)/(dx//dt) = (2//1+t^(2))/(2//1+t^(2)) = 1` [from Eqs. (iii) and (iv)]