Question #92c75

1 Answer
Jan 30, 2017

:. x=sqrt((bc)/a)[{y^((2a)/b)-(y+1)^((2a)/b)}/{y^((2a)/b)+(y+1)^((2a)/b)}]

Explanation:

{(1+xsqrt(a/(bc)))/(1-xsqrt(a/(bc)))}^(1/a)=(y/(y+1))^(2/b)

Observe that, taking a^(th) power of both sides, the power of L.H.S. will become (1/a)(a)=1, & that of the R.H.S., (2/b)(a)=(2a)/b.

{(1+xsqrt(a/(bc)))/(1-xsqrt(a/(bc)))}=(y/(y+1))^((2a)/b)=y^((2a)/b)/(y+1)^((2a)/b)

Now, we use componendo-dividendo and get,

{(1+xsqrt(a/(bc)))+(1-xsqrt(a/(bc)))}/{(1+xsqrt(a/(bc)))-(1-xsqrt(a/(bc)))}={y^((2a)/b)+(y+1)^((2a)/b)}/{y^((2a)/b)-(y+1)^((2a)/b)}

:. 1/{xsqrt(a/(bc))}={y^((2a)/b)+(y+1)^((2a)/b)}/{y^((2a)/b)-(y+1)^((2a)/b)}

:.{xsqrt(a/(bc))}={y^((2a)/b)-(y+1)^((2a)/b)}/{y^((2a)/b)+(y+1)^((2a)/b)}

:. x=sqrt((bc)/a)[{y^((2a)/b)-(y+1)^((2a)/b)}/{y^((2a)/b)+(y+1)^((2a)/b)}]

Enjoy maths.!