Hi all,

I have been running through this problem but can't get anywhere near it:

$\displaystyle x = \frac {sin(u)}{cos(v)}$ prove:

$\displaystyle \frac {\partial u}{\partial x} = \frac{cos^3(v)cos(u)}{cos(u+v)cos(u-v)} $

I start off:

$\displaystyle \frac {\partial x}{\partial u} = \frac {cos(u)}{cos(v)}=> \frac {\partial u}{\partial x} = \frac {cos(v)}{cos(u)}$ multiply top and bottom by $\displaystyle cos^2(v)cos(u)$ to give the numerator. Whats left is to prove the denominator:

$\displaystyle cos^2(u) cos^2(v) = cos(u+v)cos(u-v)$

I have tried whatever I can think of, double angle identities,product to sum and even a messy Euler exponential expansion, but fail... is there something fundamental I have missed here? Maybe this should be in the trig sub forum.....

Thanks for reading!