# Math Help - very difficult trig question

1. ## very difficult trig question

If

$\frac{ax}{\cos\theta}+\frac{by}{\sin\theta}=a^2-b^2$

and

$\frac{ax\sin\theta}{{\cos}^2\theta}-\frac{by\cos\theta}{{\sin}^2\theta}=0$

then prove that

${(ax)}^{2/3} + {(by)}^{2/3} = {(a^2 - b^2)}^{2/3}$

2. Hi

Multiply the second equation by $\frac{\sin \theta}{\cos \theta}$ and add it to the first one : $\frac{ax}{\cos^3 \theta}=a^2-b^2$

Multiply the second equation by $\frac{\cos \theta}{\sin \theta}$ and subtract it from the first one : $\frac{by}{\sin^3 \theta}=a^2-b^2$

$(ax)^{2/3} = (a^2 - b^2)^{2/3}\cos^2 \theta$
$(by)^{2/3} = (a^2 - b^2)^{2/3}\sin^2 \theta$

Then $(ax)^{2/3} + (by)^{2/3} = (a^2 - b^2)^{2/3}$