If `0ltxlt(pi)/(2)` exp `[(sin^(2)x+sin^(4)x+sin^(6)x+'.....+oo)log_(e)2]` satisfies the quadratic equation `x^(2)-9x+8=0`, find the value of `(sinx-cosx)/(sinx+cosx)`.

Explanation

`0ltxlt(pi)/(2)`
`:. 0ltsin^(2)xlt1`
Then, `sin^(2)x+sin^(4)x+sin^(6)x+"....."+oo`
`(sin^(2)x)/(1-sin^(2)x)=tan^(2)x`
`:. Exp [(sin^(2)x+sin^(4)x+sin^(6)x+'.....+oo)log_(e)2]`
`=exp(tan^(2)x*log_(e)2)=exp(log_(e)2^(tan^(2)x))`
`=e^(log_(e)tan^(2)x)=2^(tan^(2)x)`
Let `y=2^(tan^(2)x)`
Beause y satisfies the quadratic equation.
Then, `y^(2)-9y+8=0`
So, `y=1,8`
if `y=1=2^(tan^(2)x)`
`implies 2^(tan^(2)x)=2^(0)`
`implies tan^(2)x=0`
`:. x=0 ' ' [" impossible " :.xgt0]`
Now,if `y=8=2^(tan^(2)x)`
`implies 2^(tan^(2)x)=2^(3)`
`implies tan^(2)x=3`
`:. tanx= sqrt(3)`
`(sinx-cosx)/(sinx+cosx)=(tan-1)/(tan+1)=(sqrt(3)-1)/(sqrt(3)+1)xx(sqrt(3)-1)/(sqrt(3)-1)`
`=(sqrt(3)-1)^(2)/(3-1)=(3+1-2sqrt(3))/(2)`
Hence, `(sinx-cosx)/(sinx+cosx)=2-sqrt(3)`.