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. 

$a\tan\theta=b\tan\phi=c,\;$ say.  Then,
$\tan\theta=c/a, \; \tan\phi=c/b\;$ so that
$\;\sin\theta=c/\sqrt{c^2+a^2},\;  \cos\theta=a/\sqrt{c^2+a^2},\; \sin\phi=c/\sqrt{c^2+b^2}, \; \cos\phi=b/\sqrt{c^2+b^2}$.
From the first two equations, we get
$m=\dfrac{ac^2}{c^2+a^2}+\dfrac{ba^2}{c^2+b^2}, \; n=\dfrac{ab^2}{c^2+b^2}+\dfrac{bc^2}{c^2+b^2}$.
Then, $ab(m+n)=\dfrac{ab(a+b)(c^2+ab)^2}{(c^2+a^2)(c^2+b^2)}=mn(a+b)$.
