If $a\sin^2\theta+b\cos^2\theta=m, a\cos^2\phi+b\sin^2\phi=n, a\tan\theta=b\tan\phi$, prove that $\frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}$ 
If $\, a\sin^2\theta+b\cos^2\theta=m$, $\, a\cos^2\phi+b\sin^2\phi=n$, $\, a\tan\theta=b\tan\phi$.  Prove that
$$\frac{1}{n}+\frac{1}{m}=\frac{1}{a}+\frac{1}{b}$$

I've tried setting :
$$
\frac{a}{b}=\frac{\tan\theta}{\tan\phi}=k
\\ a=bk
\\
\tan\theta=k\tan\phi
$$
and substituting these in the equations. Can anyone tell me what I'm doing wrong ? A shorter path to the answer would also be appreciated. 

$$a\sin^2\theta+b\cos^2\theta=m\to a\tan^2 \theta+b=m\sec^2\theta\\
a\tan^2\theta +b=m(1+\tan^2\theta)\to \tan^2\theta=\frac{m-b}{a-m}$$
so, using the same idea
$$\tan^2\phi=\frac{n-a}{b-n}$$
we also have that 
$$\frac{\tan^2 \theta}{\tan^2 \phi}=\frac{b^2}{a^2}$$
Can you finish?
